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Speed of Mechanical Wave on String (Transverse wave) Questions in English
Class 11 Physics · Waves and Sound · Speed of Mechanical Wave on String (Transverse wave)
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51
DifficultMCQ
The mass per unit length of a uniform wire is $0.135 \, g/cm$. $A$ transverse wave of the form $y = -0.21 \sin(x + 30t)$ is produced in it,where $x$ is in meters and $t$ is in seconds. Then,the expected value of tension in the wire is $x \times 10^{-2} \, N$. The value of $x$ is (Round off to the nearest integer).
A
$12.15$
B
$121.5$
C
$1215$
D
$24.3$
Solution
(C) Given mass per unit length $\mu = 0.135 \, g/cm = 0.135 \times 10^{-3} \, kg / 10^{-2} \, m = 0.0135 \, kg/m$.
The wave equation is $y = -0.21 \sin(x + 30t)$.
Comparing this with the standard wave equation $y = A \sin(kx + \omega t)$,we get angular frequency $\omega = 30 \, rad/s$ and wave number $k = 1 \, m^{-1}$.
The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{30}{1} = 30 \, m/s$.
The speed of a transverse wave on a stretched string is $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension.
Therefore,$T = v^2 \mu = (30)^2 \times 0.0135 = 900 \times 0.0135 = 12.15 \, N$.
We are given $T = x \times 10^{-2} \, N$,so $12.15 = x \times 10^{-2}$.
Thus,$x = 12.15 \times 100 = 1215$.
The wave equation is $y = -0.21 \sin(x + 30t)$.
Comparing this with the standard wave equation $y = A \sin(kx + \omega t)$,we get angular frequency $\omega = 30 \, rad/s$ and wave number $k = 1 \, m^{-1}$.
The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{30}{1} = 30 \, m/s$.
The speed of a transverse wave on a stretched string is $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension.
Therefore,$T = v^2 \mu = (30)^2 \times 0.0135 = 900 \times 0.0135 = 12.15 \, N$.
We are given $T = x \times 10^{-2} \, N$,so $12.15 = x \times 10^{-2}$.
Thus,$x = 12.15 \times 100 = 1215$.
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View Solution52
MediumMCQ
The percentage increase in the speed of transverse waves produced in a stretched string if the tension is increased by $4\, \%$,will be ......... $\%$
A
$1$
B
$2$
C
$0$
D
$4$
Solution
(B) The speed of a transverse wave in a stretched string is given by the formula $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Taking the natural logarithm on both sides,we get $\ln v = \frac{1}{2} \ln T - \frac{1}{2} \ln \mu$.
Differentiating both sides,we obtain the relative error formula: $\frac{\Delta v}{v} = \frac{1}{2} \frac{\Delta T}{T}$.
Given that the percentage increase in tension is $\frac{\Delta T}{T} \times 100 = 4\, \%$.
Substituting this value into the error formula: $\frac{\Delta v}{v} \times 100 = \frac{1}{2} \times (4\, \%) = 2\, \%$.
Therefore,the percentage increase in the speed of the wave is $2\, \%$.
Taking the natural logarithm on both sides,we get $\ln v = \frac{1}{2} \ln T - \frac{1}{2} \ln \mu$.
Differentiating both sides,we obtain the relative error formula: $\frac{\Delta v}{v} = \frac{1}{2} \frac{\Delta T}{T}$.
Given that the percentage increase in tension is $\frac{\Delta T}{T} \times 100 = 4\, \%$.
Substituting this value into the error formula: $\frac{\Delta v}{v} \times 100 = \frac{1}{2} \times (4\, \%) = 2\, \%$.
Therefore,the percentage increase in the speed of the wave is $2\, \%$.
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View Solution53
EasyMCQ
If the initial tension on a stretched string is doubled,then the ratio of the initial and final speeds of a transverse wave along the string is:
A
$ \sqrt{2} : 1 $
B
$ 1 : \sqrt{2} $
C
$ 1 : 2 $
D
$ 1 : 1 $
Solution
(B) The speed of a transverse wave on a stretched string is given by the formula $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Since $\mu$ remains constant,the speed is directly proportional to the square root of the tension: $v \propto \sqrt{T}$.
Let the initial tension be $T_i = T$ and the final tension be $T_f = 2T$.
The ratio of the initial speed $v_i$ to the final speed $v_f$ is given by:
$\frac{v_i}{v_f} = \sqrt{\frac{T_i}{T_f}}$
Substituting the values,we get:
$\frac{v_i}{v_f} = \sqrt{\frac{T}{2T}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$
Thus,the ratio is $1 : \sqrt{2}$.
Since $\mu$ remains constant,the speed is directly proportional to the square root of the tension: $v \propto \sqrt{T}$.
Let the initial tension be $T_i = T$ and the final tension be $T_f = 2T$.
The ratio of the initial speed $v_i$ to the final speed $v_f$ is given by:
$\frac{v_i}{v_f} = \sqrt{\frac{T_i}{T_f}}$
Substituting the values,we get:
$\frac{v_i}{v_f} = \sqrt{\frac{T}{2T}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$
Thus,the ratio is $1 : \sqrt{2}$.
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View Solution54
AdvancedMCQ
$A$ rope of length $L$ and uniform linear density is hanging from the ceiling. $A$ transverse wave pulse,generated close to the free end of the rope,travels upwards through the rope. Select the correct option.
A
The speed of the pulse decreases as it moves up.
B
The time taken by the pulse to travel the length of the rope is proportional to $\sqrt{L}$.
C
The tension will be constant along the length of the rope.
D
The speed of the pulse will be constant along the length of the rope.
Solution
(B) The tension $T$ at a section at a distance $x$ from the free end is given by $T = m g = \mu x g$,where $\mu$ is the mass per unit length of the rope.
The wave speed $v$ on the rope is given by $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{\mu x g}{\mu}} = \sqrt{g x}$.
Since $v = \sqrt{g x}$,the speed increases as the pulse moves up (as $x$ increases).
To find the time $t$ taken to travel a distance $x$,we use $v = \frac{dx}{dt} = \sqrt{g x}$.
Rearranging gives $dx / \sqrt{x} = \sqrt{g} dt$.
Integrating both sides from $0$ to $x$ and $0$ to $t$,we get $\int_{0}^{x} x^{-1/2} dx = \int_{0}^{t} \sqrt{g} dt$.
$2\sqrt{x} = \sqrt{g} t$,which implies $t = 2\sqrt{\frac{x}{g}}$.
For the total length $L$,the time taken is $t = 2\sqrt{\frac{L}{g}}$,so $t \propto \sqrt{L}$.
The wave speed $v$ on the rope is given by $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{\mu x g}{\mu}} = \sqrt{g x}$.
Since $v = \sqrt{g x}$,the speed increases as the pulse moves up (as $x$ increases).
To find the time $t$ taken to travel a distance $x$,we use $v = \frac{dx}{dt} = \sqrt{g x}$.
Rearranging gives $dx / \sqrt{x} = \sqrt{g} dt$.
Integrating both sides from $0$ to $x$ and $0$ to $t$,we get $\int_{0}^{x} x^{-1/2} dx = \int_{0}^{t} \sqrt{g} dt$.
$2\sqrt{x} = \sqrt{g} t$,which implies $t = 2\sqrt{\frac{x}{g}}$.
For the total length $L$,the time taken is $t = 2\sqrt{\frac{L}{g}}$,so $t \propto \sqrt{L}$.
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View Solution55
MediumMCQ
$A$ transverse wave propagating on a string can be described by the equation $y = 2 \sin (10x + 300t)$,where $x$ and $y$ are in meters and $t$ is in seconds. If the vibrating string has a linear mass density of $0.6 \times 10^{-3} \, g/cm$,then the tension in the string is .............. $N$.
A
$5.4$
B
$0.054$
C
$54$
D
$0.0054$
Solution
(B) The given wave equation is $y = 2 \sin (10x + 300t)$.
Comparing this with the standard wave equation $y = A \sin (kx + \omega t)$,we get the wave number $k = 10 \, rad/m$ and the angular frequency $\omega = 300 \, rad/s$.
The linear mass density $\mu = 0.6 \times 10^{-3} \, g/cm$. Converting this to $SI$ units $(kg/m)$:
$\mu = \frac{0.6 \times 10^{-3} \times 10^{-3} \, kg}{10^{-2} \, m} = 6 \times 10^{-5} \, kg/m$.
The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{300}{10} = 30 \, m/s$.
The speed of a transverse wave on a string is related to tension $T$ and linear mass density $\mu$ by $v = \sqrt{\frac{T}{\mu}}$.
Therefore,$T = v^2 \mu$.
Substituting the values: $T = (30)^2 \times (6 \times 10^{-5}) = 900 \times 6 \times 10^{-5} = 5400 \times 10^{-5} = 0.054 \, N$.
Comparing this with the standard wave equation $y = A \sin (kx + \omega t)$,we get the wave number $k = 10 \, rad/m$ and the angular frequency $\omega = 300 \, rad/s$.
The linear mass density $\mu = 0.6 \times 10^{-3} \, g/cm$. Converting this to $SI$ units $(kg/m)$:
$\mu = \frac{0.6 \times 10^{-3} \times 10^{-3} \, kg}{10^{-2} \, m} = 6 \times 10^{-5} \, kg/m$.
The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{300}{10} = 30 \, m/s$.
The speed of a transverse wave on a string is related to tension $T$ and linear mass density $\mu$ by $v = \sqrt{\frac{T}{\mu}}$.
Therefore,$T = v^2 \mu$.
Substituting the values: $T = (30)^2 \times (6 \times 10^{-5}) = 900 \times 6 \times 10^{-5} = 5400 \times 10^{-5} = 0.054 \, N$.
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View Solution56
DifficultMCQ
$A$ copper wire is held at the two ends by rigid supports. At $50^{\circ} C$ the wire is just taut,with negligible tension. If $Y=1.2 \times 10^{11} \, N/m^2$,$\alpha=1.6 \times 10^{-5} /^{\circ} C$,and $\rho=9.2 \times 10^3 \, kg/m^3$,then the speed of transverse waves in this wire at $30^{\circ} C$ is .......... $m/s$.
A
$64.6$
B
$16.2$
C
$23.2$
D
$32.2$
Solution
(A) The thermal stress $F/A$ developed in a wire due to a temperature change $\Delta T$ is given by $F/A = Y \alpha \Delta T$.
The speed of a transverse wave on a stretched wire is $v = \sqrt{T/\mu}$,where $T$ is the tension and $\mu$ is the linear mass density.
Here,the tension $T = F = Y A \alpha \Delta T$ and the linear mass density $\mu = \rho A$.
Substituting these into the velocity formula: $v = \sqrt{\frac{Y A \alpha \Delta T}{\rho A}} = \sqrt{\frac{Y \alpha \Delta T}{\rho}}$.
Given: $Y = 1.2 \times 10^{11} \, N/m^2$,$\alpha = 1.6 \times 10^{-5} /^{\circ} C$,$\rho = 9.2 \times 10^3 \, kg/m^3$,and $\Delta T = 50^{\circ} C - 30^{\circ} C = 20^{\circ} C$.
Calculating the speed: $v = \sqrt{\frac{1.2 \times 10^{11} \times 1.6 \times 10^{-5} \times 20}{9.2 \times 10^3}}$.
$v = \sqrt{\frac{3.84 \times 10^7}{9.2 \times 10^3}} = \sqrt{4173.9} \approx 64.6 \, m/s$.
The speed of a transverse wave on a stretched wire is $v = \sqrt{T/\mu}$,where $T$ is the tension and $\mu$ is the linear mass density.
Here,the tension $T = F = Y A \alpha \Delta T$ and the linear mass density $\mu = \rho A$.
Substituting these into the velocity formula: $v = \sqrt{\frac{Y A \alpha \Delta T}{\rho A}} = \sqrt{\frac{Y \alpha \Delta T}{\rho}}$.
Given: $Y = 1.2 \times 10^{11} \, N/m^2$,$\alpha = 1.6 \times 10^{-5} /^{\circ} C$,$\rho = 9.2 \times 10^3 \, kg/m^3$,and $\Delta T = 50^{\circ} C - 30^{\circ} C = 20^{\circ} C$.
Calculating the speed: $v = \sqrt{\frac{1.2 \times 10^{11} \times 1.6 \times 10^{-5} \times 20}{9.2 \times 10^3}}$.
$v = \sqrt{\frac{3.84 \times 10^7}{9.2 \times 10^3}} = \sqrt{4173.9} \approx 64.6 \, m/s$.
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View Solution57
MediumMCQ
$A$ pulse is generated at the lower end of a hanging rope of uniform density and length $L$. The speed of the pulse when it reaches the midpoint of the rope is:
A
$\sqrt{2 g L}$
B
$\sqrt{g L}$
C
$\sqrt{\frac{g L}{2}}$
D
$\frac{\sqrt{g L}}{2}$
Solution
(C) The speed of a transverse wave on a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density of the string.
For a hanging rope of length $L$ and linear mass density $\mu$,the tension $T$ at a distance $y$ from the lower end is due to the weight of the rope segment below it.
The mass of the rope segment of length $y$ is $m = \mu y$.
Thus,the tension at distance $y$ is $T = \mu y g$.
The speed of the pulse at distance $y$ is $v = \sqrt{\frac{\mu y g}{\mu}} = \sqrt{gy}$.
At the midpoint of the rope,$y = \frac{L}{2}$.
Therefore,the speed of the pulse at the midpoint is $v = \sqrt{g \left(\frac{L}{2}\right)} = \sqrt{\frac{g L}{2}}$.
For a hanging rope of length $L$ and linear mass density $\mu$,the tension $T$ at a distance $y$ from the lower end is due to the weight of the rope segment below it.
The mass of the rope segment of length $y$ is $m = \mu y$.
Thus,the tension at distance $y$ is $T = \mu y g$.
The speed of the pulse at distance $y$ is $v = \sqrt{\frac{\mu y g}{\mu}} = \sqrt{gy}$.
At the midpoint of the rope,$y = \frac{L}{2}$.
Therefore,the speed of the pulse at the midpoint is $v = \sqrt{g \left(\frac{L}{2}\right)} = \sqrt{\frac{g L}{2}}$.
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View Solution58
DifficultMCQ
$A$ rope of length $L$ and mass $M$ hangs freely from the ceiling. If the time taken by a transverse wave to travel from the bottom to the top of the rope is $T$,then the time taken to cover the first half length is:
A
$T$
B
$T\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)$
C
$\frac{T}{\sqrt{2}}$
D
$\frac{T}{2}$
Solution
(C) The speed of a transverse wave on a string is given by $v = \sqrt{\frac{T_s}{\mu}}$,where $T_s$ is the tension and $\mu = \frac{M}{L}$ is the mass per unit length.
At a distance $x$ from the bottom,the tension $T_s$ is due to the mass of the rope below it,which is $m = \frac{M}{L} x$. Thus,$T_s = \left(\frac{M}{L} x\right)g$.
The speed at distance $x$ is $v(x) = \sqrt{\frac{(M/L)xg}{M/L}} = \sqrt{gx}$.
Since $v = \frac{dx}{dt}$,we have $dt = \frac{dx}{\sqrt{gx}}$.
Integrating from $x=0$ to $x=L$ to find the total time $T$:
$T = \int_0^L \frac{dx}{\sqrt{gx}} = \frac{1}{\sqrt{g}} [2\sqrt{x}]_0^L = 2\sqrt{\frac{L}{g}}$.
Now,to find the time $T_1$ to cover the first half length ($x=0$ to $x=L/2$):
$T_1 = \int_0^{L/2} \frac{dx}{\sqrt{gx}} = \frac{1}{\sqrt{g}} [2\sqrt{x}]_0^{L/2} = 2\sqrt{\frac{L}{2g}} = \frac{2}{\sqrt{2}}\sqrt{\frac{L}{g}} = \frac{T}{\sqrt{2}}$.
At a distance $x$ from the bottom,the tension $T_s$ is due to the mass of the rope below it,which is $m = \frac{M}{L} x$. Thus,$T_s = \left(\frac{M}{L} x\right)g$.
The speed at distance $x$ is $v(x) = \sqrt{\frac{(M/L)xg}{M/L}} = \sqrt{gx}$.
Since $v = \frac{dx}{dt}$,we have $dt = \frac{dx}{\sqrt{gx}}$.
Integrating from $x=0$ to $x=L$ to find the total time $T$:
$T = \int_0^L \frac{dx}{\sqrt{gx}} = \frac{1}{\sqrt{g}} [2\sqrt{x}]_0^L = 2\sqrt{\frac{L}{g}}$.
Now,to find the time $T_1$ to cover the first half length ($x=0$ to $x=L/2$):
$T_1 = \int_0^{L/2} \frac{dx}{\sqrt{gx}} = \frac{1}{\sqrt{g}} [2\sqrt{x}]_0^{L/2} = 2\sqrt{\frac{L}{2g}} = \frac{2}{\sqrt{2}}\sqrt{\frac{L}{g}} = \frac{T}{\sqrt{2}}$.
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View Solution59
DifficultMCQ
$A$ transverse pulse generated at the bottom of a uniform rope of length $L$ travels in an upward direction. The time taken by it to travel the full length of the rope will be
A
$\sqrt{\frac{L}{2g}}$
B
$\sqrt{\frac{2L}{g}}$
C
$\sqrt{\frac{L}{g}}$
D
$\sqrt{\frac{4L}{g}}$
Solution
(D) The velocity of a transverse wave on a string at a distance $x$ from the free end is given by $v = \sqrt{gx}$.
Since the pulse starts from the bottom (free end),at a distance $x$ from the bottom,the velocity is $v = \frac{dx}{dt} = \sqrt{gx}$.
Rearranging for time,we get $dt = \frac{dx}{\sqrt{gx}}$.
To find the total time $T$ to travel the length $L$,we integrate from $0$ to $L$:
$T = \int_{0}^{L} \frac{dx}{\sqrt{gx}} = \frac{1}{\sqrt{g}} \int_{0}^{L} x^{-1/2} dx$.
$T = \frac{1}{\sqrt{g}} [2x^{1/2}]_{0}^{L} = \frac{1}{\sqrt{g}} (2\sqrt{L}) = 2\sqrt{\frac{L}{g}} = \sqrt{\frac{4L}{g}}$.
Thus,the correct option is $D$.
Since the pulse starts from the bottom (free end),at a distance $x$ from the bottom,the velocity is $v = \frac{dx}{dt} = \sqrt{gx}$.
Rearranging for time,we get $dt = \frac{dx}{\sqrt{gx}}$.
To find the total time $T$ to travel the length $L$,we integrate from $0$ to $L$:
$T = \int_{0}^{L} \frac{dx}{\sqrt{gx}} = \frac{1}{\sqrt{g}} \int_{0}^{L} x^{-1/2} dx$.
$T = \frac{1}{\sqrt{g}} [2x^{1/2}]_{0}^{L} = \frac{1}{\sqrt{g}} (2\sqrt{L}) = 2\sqrt{\frac{L}{g}} = \sqrt{\frac{4L}{g}}$.
Thus,the correct option is $D$.
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View Solution60
EasyMCQ
$A$ steel wire with mass per unit length $7.0 \times 10^{-3} \, kg \, m^{-1}$ is under tension of $70 \, N$. The speed of transverse waves in the wire will be $......... \, m/s$.
A
$200$
B
$100$
C
$10$
D
$50$
Solution
(B) The speed of a transverse wave on a stretched string is given by the formula $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the mass per unit length.
Given:
Tension $T = 70 \, N$
Mass per unit length $\mu = 7.0 \times 10^{-3} \, kg \, m^{-1}$
Substituting the values into the formula:
$v = \sqrt{\frac{70}{7.0 \times 10^{-3}}}$
$v = \sqrt{\frac{70}{0.007}}$
$v = \sqrt{10000}$
$v = 100 \, m/s$
Therefore,the speed of the transverse wave is $100 \, m/s$.
Given:
Tension $T = 70 \, N$
Mass per unit length $\mu = 7.0 \times 10^{-3} \, kg \, m^{-1}$
Substituting the values into the formula:
$v = \sqrt{\frac{70}{7.0 \times 10^{-3}}}$
$v = \sqrt{\frac{70}{0.007}}$
$v = \sqrt{10000}$
$v = 100 \, m/s$
Therefore,the speed of the transverse wave is $100 \, m/s$.
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View Solution61
MediumMCQ
The fundamental frequency of vibration of a string stretched between two rigid supports is $50\,Hz$. The mass of the string is $18\,g$ and its linear mass density is $20\,g/m$. The speed of the transverse waves produced in the string is $..........\,m/s$.
A
$90$
B
$45$
C
$30$
D
$15$
Solution
(A) Given:
Fundamental frequency $f = 50\,Hz$
Mass of the string $m = 18\,g = 0.018\,kg$
Linear mass density $\mu = 20\,g/m = 0.02\,kg/m$
Step $1$: Find the length of the string $(L)$.
We know that $\mu = \frac{m}{L}$,so $L = \frac{m}{\mu} = \frac{18\,g}{20\,g/m} = 0.9\,m$.
Step $2$: Relate frequency to wave speed.
For the fundamental mode of a string fixed at both ends,the length $L$ is equal to half the wavelength: $L = \frac{\lambda}{2}$,which implies $\lambda = 2L$.
Substituting $L = 0.9\,m$,we get $\lambda = 2 \times 0.9 = 1.8\,m$.
Step $3$: Calculate the wave speed $(v)$.
The speed of the wave is given by $v = f \lambda$.
$v = 50\,Hz \times 1.8\,m = 90\,m/s$.
Thus,the speed of the transverse waves is $90\,m/s$.
Fundamental frequency $f = 50\,Hz$
Mass of the string $m = 18\,g = 0.018\,kg$
Linear mass density $\mu = 20\,g/m = 0.02\,kg/m$
Step $1$: Find the length of the string $(L)$.
We know that $\mu = \frac{m}{L}$,so $L = \frac{m}{\mu} = \frac{18\,g}{20\,g/m} = 0.9\,m$.
Step $2$: Relate frequency to wave speed.
For the fundamental mode of a string fixed at both ends,the length $L$ is equal to half the wavelength: $L = \frac{\lambda}{2}$,which implies $\lambda = 2L$.
Substituting $L = 0.9\,m$,we get $\lambda = 2 \times 0.9 = 1.8\,m$.
Step $3$: Calculate the wave speed $(v)$.
The speed of the wave is given by $v = f \lambda$.
$v = 50\,Hz \times 1.8\,m = 90\,m/s$.
Thus,the speed of the transverse waves is $90\,m/s$.
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View Solution62
AdvancedMCQ
$A$ block $M$ hangs vertically at the bottom end of a uniform rope of constant mass per unit length. The top end of the rope is attached to a fixed rigid support at $O$. $A$ transverse wave pulse (Pulse $1$) of wavelength $\lambda_0$ is produced at point $O$ on the rope. The pulse takes time $T_{OA}$ to reach point $A$. If the wave pulse of wavelength $\lambda_0$ is produced at point $A$ (Pulse $2$) without disturbing the position of $M$,it takes time $T_{AO}$ to reach point $O$. Which of the following options is/are correct?
A
$B, C, D$
B
$A, B, D$
C
$B, C$
D
$C, D$
Solution
(B) The speed of a transverse pulse at any point on the rope is given by $v = \sqrt{\frac{T(x)}{\mu}}$,where $T(x)$ is the tension at that point and $\mu$ is the linear mass density.
$1$. For Pulse $1$ moving from $O$ to $A$,the tension at a distance $x$ from $O$ is $T(x) = (M + \mu x)g$.
$2$. For Pulse $2$ moving from $A$ to $O$,the tension at a distance $x$ from $O$ is also $T(x) = (M + \mu x)g$.
Since the tension profile along the rope is identical for both paths,the speed at any given point $x$ is the same for both pulses. Thus,the time taken $T_{OA} = T_{AO}$. Option $A$ is correct.
$3$. Since the speed $v(x)$ is the same at any point $x$ for both pulses,option $B$ is correct.
$4$. The speed of a transverse wave on a string is $v = \sqrt{T/\mu}$,which is independent of frequency and wavelength. Thus,option $D$ is correct.
$5$. As the pulse moves from $O$ to $A$,the tension increases,so the speed $v$ increases. Since $v = f\lambda$ and the frequency $f$ remains constant,the wavelength $\lambda$ must increase. Thus,option $C$ is correct.
Therefore,options $A, B, C,$ and $D$ are all correct.
$1$. For Pulse $1$ moving from $O$ to $A$,the tension at a distance $x$ from $O$ is $T(x) = (M + \mu x)g$.
$2$. For Pulse $2$ moving from $A$ to $O$,the tension at a distance $x$ from $O$ is also $T(x) = (M + \mu x)g$.
Since the tension profile along the rope is identical for both paths,the speed at any given point $x$ is the same for both pulses. Thus,the time taken $T_{OA} = T_{AO}$. Option $A$ is correct.
$3$. Since the speed $v(x)$ is the same at any point $x$ for both pulses,option $B$ is correct.
$4$. The speed of a transverse wave on a string is $v = \sqrt{T/\mu}$,which is independent of frequency and wavelength. Thus,option $D$ is correct.
$5$. As the pulse moves from $O$ to $A$,the tension increases,so the speed $v$ increases. Since $v = f\lambda$ and the frequency $f$ remains constant,the wavelength $\lambda$ must increase. Thus,option $C$ is correct.
Therefore,options $A, B, C,$ and $D$ are all correct.
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View Solution63
MediumMCQ
Two strings with circular cross section and made of same material are stretched to have the same amount of tension. $A$ transverse wave is then made to pass through both the strings. The velocity of the wave in the first string having the radius of cross section $R$ is $v_1$,and that in the other string having radius of cross section $R/2$ is $v_2$. Then $\frac{v_2}{v_1} = $
A
$\sqrt{2}$
B
$2$
C
$8$
D
$4$
Solution
(B) The velocity of a transverse wave in a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Linear mass density $\mu = \rho A = \rho (\pi R^2)$,where $\rho$ is the density of the material and $R$ is the radius of the cross section.
Since both strings are made of the same material,$\rho$ is constant. Given that the tension $T$ is the same for both strings,we have $v \propto \frac{1}{\sqrt{R^2}} \propto \frac{1}{R}$.
Therefore,$\frac{v_2}{v_1} = \frac{R_1}{R_2}$.
Given $R_1 = R$ and $R_2 = R/2$,we get $\frac{v_2}{v_1} = \frac{R}{R/2} = 2$.
Linear mass density $\mu = \rho A = \rho (\pi R^2)$,where $\rho$ is the density of the material and $R$ is the radius of the cross section.
Since both strings are made of the same material,$\rho$ is constant. Given that the tension $T$ is the same for both strings,we have $v \propto \frac{1}{\sqrt{R^2}} \propto \frac{1}{R}$.
Therefore,$\frac{v_2}{v_1} = \frac{R_1}{R_2}$.
Given $R_1 = R$ and $R_2 = R/2$,we get $\frac{v_2}{v_1} = \frac{R}{R/2} = 2$.
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View Solution64
MediumMCQ
The linear mass density of a vibrating string is $1.3 \times 10^{-4} \ kg/m$. $A$ transverse wave is propagating on the string and is described by the equation $y = 0.021 \sin(x + 30t)$,where $x$ and $y$ are measured in meters and $t$ in seconds. The tension in the string is approximately: (in $N$)
A
$0.12$
B
$0.48$
C
$1.20$
D
$4.80$
Solution
(A) The standard equation of a transverse wave is $y = A \sin(kx + \omega t)$.
Comparing this with the given equation $y = 0.021 \sin(x + 30t)$,we get the angular frequency $\omega = 30 \ rad/s$ and the wave number $k = 1 \ m^{-1}$.
The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{30}{1} = 30 \ m/s$.
The tension $T$ in a string is related to the linear mass density $\mu$ and wave speed $v$ by the formula $T = \mu v^2$.
Given $\mu = 1.3 \times 10^{-4} \ kg/m$,we have $T = (1.3 \times 10^{-4}) \times (30)^2$.
$T = 1.3 \times 10^{-4} \times 900 = 1.3 \times 0.09 = 0.117 \ N$.
Rounding to two decimal places,the tension is approximately $0.12 \ N$.
Comparing this with the given equation $y = 0.021 \sin(x + 30t)$,we get the angular frequency $\omega = 30 \ rad/s$ and the wave number $k = 1 \ m^{-1}$.
The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{30}{1} = 30 \ m/s$.
The tension $T$ in a string is related to the linear mass density $\mu$ and wave speed $v$ by the formula $T = \mu v^2$.
Given $\mu = 1.3 \times 10^{-4} \ kg/m$,we have $T = (1.3 \times 10^{-4}) \times (30)^2$.
$T = 1.3 \times 10^{-4} \times 900 = 1.3 \times 0.09 = 0.117 \ N$.
Rounding to two decimal places,the tension is approximately $0.12 \ N$.
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View Solution65
AdvancedMCQ
Consider a system of three connected strings,$S_1, S_2$ and $S_3$ with uniform linear mass densities $\mu \text{ kg/m}$,$4\mu \text{ kg/m}$ and $16\mu \text{ kg/m}$,respectively,as shown in the figure. $S_1$ and $S_2$ are connected at the point $P$,whereas $S_2$ and $S_3$ are connected at the point $Q$,and the other end of $S_3$ is connected to a wall. $A$ wave generator $O$ is connected to the free end of $S_1$. The wave from the generator is represented by $y = y_0 \cos(\omega t - kx) \text{ cm}$,where $y_0, \omega$ and $k$ are constants of appropriate dimensions. Which of the following statements is/are correct:
$(A)$ When the wave reflects from $P$ for the first time,the reflected wave is represented by $y = \alpha_1 y_0 \cos(\omega t + kx + \pi) \text{ cm}$,where $\alpha_1$ is a positive constant.
$(B)$ When the wave transmits through $P$ for the first time,the transmitted wave is represented by $y = \alpha_2 y_0 \cos(\omega t - kx) \text{ cm}$,where $\alpha_2$ is a positive constant.
$(C)$ When the wave reflects from $Q$ for the first time,the reflected wave is represented by $y = \alpha_3 y_0 \cos(\omega t - kx + \pi) \text{ cm}$,where $\alpha_3$ is a positive constant.
$(D)$ When the wave transmits through $Q$ for the first time,the transmitted wave is represented by $y = \alpha_4 y_0 \cos(\omega t - 4kx) \text{ cm}$,where $\alpha_4$ is a positive constant.
$(A)$ When the wave reflects from $P$ for the first time,the reflected wave is represented by $y = \alpha_1 y_0 \cos(\omega t + kx + \pi) \text{ cm}$,where $\alpha_1$ is a positive constant.
$(B)$ When the wave transmits through $P$ for the first time,the transmitted wave is represented by $y = \alpha_2 y_0 \cos(\omega t - kx) \text{ cm}$,where $\alpha_2$ is a positive constant.
$(C)$ When the wave reflects from $Q$ for the first time,the reflected wave is represented by $y = \alpha_3 y_0 \cos(\omega t - kx + \pi) \text{ cm}$,where $\alpha_3$ is a positive constant.
$(D)$ When the wave transmits through $Q$ for the first time,the transmitted wave is represented by $y = \alpha_4 y_0 \cos(\omega t - 4kx) \text{ cm}$,where $\alpha_4$ is a positive constant.
A
$(A, C)$
B
$(A, D)$
C
$(B, C)$
D
$(B, D)$
Solution
(B) The incident wave is $y_i = y_0 \cos(\omega t - kx)$.
At point $P$,the wave travels from a medium with density $\mu$ to $4\mu$. Since it moves from a rarer to a denser medium,the reflected wave undergoes a phase change of $\pi$. Thus,$y_r = \alpha_1 y_0 \cos(\omega t + kx + \pi)$. Statement $(A)$ is correct.
The transmitted wave in $S_2$ has the same frequency $\omega$. The wave speed $v = \sqrt{T/\mu}$. Since $T$ is constant,$v \propto 1/\sqrt{\mu}$. Thus,$v_2 = v_1 / \sqrt{4} = v_1 / 2$. Since $v = \omega/k$,we have $k_2 = 2k$. The transmitted wave is $y_t = \alpha_2 y_0 \cos(\omega t - 2kx)$. Statement $(B)$ is incorrect.
At point $Q$,the wave travels from $S_2$ (density $4\mu$) to $S_3$ (density $16\mu$). The incident wave on $Q$ is $y_i = \alpha_2 y_0 \cos(\omega t - 2kx)$. Since it moves from a rarer to a denser medium,the reflected wave at $Q$ undergoes a phase change of $\pi$. Thus,$y_r = \alpha_3 y_0 \cos(\omega t + 2kx + \pi)$. Statement $(C)$ is incorrect.
For the transmitted wave in $S_3$,$v_3 = v_2 / \sqrt{16/4} = v_2 / 2 = v_1 / 4$. Thus,$k_3 = 4k$. The transmitted wave is $y_t = \alpha_4 y_0 \cos(\omega t - 4kx)$. Statement $(D)$ is correct.
Therefore,the correct statements are $(A)$ and $(D)$.
At point $P$,the wave travels from a medium with density $\mu$ to $4\mu$. Since it moves from a rarer to a denser medium,the reflected wave undergoes a phase change of $\pi$. Thus,$y_r = \alpha_1 y_0 \cos(\omega t + kx + \pi)$. Statement $(A)$ is correct.
The transmitted wave in $S_2$ has the same frequency $\omega$. The wave speed $v = \sqrt{T/\mu}$. Since $T$ is constant,$v \propto 1/\sqrt{\mu}$. Thus,$v_2 = v_1 / \sqrt{4} = v_1 / 2$. Since $v = \omega/k$,we have $k_2 = 2k$. The transmitted wave is $y_t = \alpha_2 y_0 \cos(\omega t - 2kx)$. Statement $(B)$ is incorrect.
At point $Q$,the wave travels from $S_2$ (density $4\mu$) to $S_3$ (density $16\mu$). The incident wave on $Q$ is $y_i = \alpha_2 y_0 \cos(\omega t - 2kx)$. Since it moves from a rarer to a denser medium,the reflected wave at $Q$ undergoes a phase change of $\pi$. Thus,$y_r = \alpha_3 y_0 \cos(\omega t + 2kx + \pi)$. Statement $(C)$ is incorrect.
For the transmitted wave in $S_3$,$v_3 = v_2 / \sqrt{16/4} = v_2 / 2 = v_1 / 4$. Thus,$k_3 = 4k$. The transmitted wave is $y_t = \alpha_4 y_0 \cos(\omega t - 4kx)$. Statement $(D)$ is correct.
Therefore,the correct statements are $(A)$ and $(D)$.
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View Solution66
DifficultMCQ
$A$ transverse wave is travelling with velocity $V$ through a metal wire of length $L$ and density $\rho$. The tensile stress in the wire is
A
$V \rho^{2}$
B
$\frac{V^{2}}{\rho}$
C
$\frac{\rho}{V^{2}}$
D
$V^{2} \rho$
Solution
(D) The velocity of a transverse wave in a stretched string is given by $V = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the mass per unit length (linear mass density).
$\mu = \frac{M}{L} = \frac{A \cdot L \cdot \rho}{L} = A \cdot \rho$,where $A$ is the cross-sectional area and $\rho$ is the material density.
Substituting $\mu$ into the velocity equation: $V = \sqrt{\frac{T}{A \cdot \rho}}$.
Squaring both sides: $V^{2} = \frac{T}{A \cdot \rho}$.
Rearranging to find the tensile stress (which is defined as $\text{Stress} = \frac{T}{A}$): $\frac{T}{A} = V^{2} \cdot \rho$.
Therefore,the tensile stress in the wire is $V^{2} \rho$.
$\mu = \frac{M}{L} = \frac{A \cdot L \cdot \rho}{L} = A \cdot \rho$,where $A$ is the cross-sectional area and $\rho$ is the material density.
Substituting $\mu$ into the velocity equation: $V = \sqrt{\frac{T}{A \cdot \rho}}$.
Squaring both sides: $V^{2} = \frac{T}{A \cdot \rho}$.
Rearranging to find the tensile stress (which is defined as $\text{Stress} = \frac{T}{A}$): $\frac{T}{A} = V^{2} \cdot \rho$.
Therefore,the tensile stress in the wire is $V^{2} \rho$.
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View Solution67
DifficultMCQ
$A$ string has a mass per unit length of $10^{-6} \,kg/cm$. The equation of a simple harmonic wave produced in it is $Y=0.2 \sin(2x+80t) \,m$. The tension in the string is: (in $N$)
A
$0.16$
B
$0.0016$
C
$1.6$
D
$16$
Solution
(A) The wave velocity $v$ is given by the formula $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the mass per unit length.
Rearranging for tension,we get $T = v^2 \mu$.
Given $\mu = 10^{-6} \,kg/cm = 10^{-4} \,kg/m$.
Comparing the given wave equation $Y = 0.2 \sin(2x + 80t)$ with the standard form $Y = A \sin(Kx + \omega t)$,we identify the angular frequency $\omega = 80 \,rad/s$ and the wave number $K = 2 \,rad/m$.
The wave velocity is $v = \frac{\omega}{K} = \frac{80}{2} = 40 \,m/s$.
Substituting these values into the tension formula:
$T = (40)^2 \times 10^{-4} = 1600 \times 10^{-4} = 0.16 \,N$.
Rearranging for tension,we get $T = v^2 \mu$.
Given $\mu = 10^{-6} \,kg/cm = 10^{-4} \,kg/m$.
Comparing the given wave equation $Y = 0.2 \sin(2x + 80t)$ with the standard form $Y = A \sin(Kx + \omega t)$,we identify the angular frequency $\omega = 80 \,rad/s$ and the wave number $K = 2 \,rad/m$.
The wave velocity is $v = \frac{\omega}{K} = \frac{80}{2} = 40 \,m/s$.
Substituting these values into the tension formula:
$T = (40)^2 \times 10^{-4} = 1600 \times 10^{-4} = 0.16 \,N$.
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View Solution68
MediumMCQ
$A$ wire $PQ$ has length $4.8 \ m$ and mass $0.06 \ kg$. Another wire $QR$ has length $2.56 \ m$ and mass $0.2 \ kg$. Both wires have the same radii and are joined as a single wire. This wire is under a tension of $80 \ N$. $A$ wave pulse of amplitude $3.5 \ cm$ is sent along the wire $PQ$ from end $P$. The time taken by the wave to reach the other end of the single wire is (No power is dissipated during propagation). (in $s$)
A
$0.1$
B
$0.12$
C
$0.14$
D
$0.16$
Solution
(C) Mass per unit length for wire $PQ$ is $m_{PQ} = \frac{0.06}{4.8} = \frac{1}{80} \ kg/m$.
Mass per unit length for wire $QR$ is $m_{QR} = \frac{0.2}{2.56} = \frac{5}{64} \ kg/m$.
The velocity of the wave in wire $PQ$ is $v_{PQ} = \sqrt{\frac{T}{m_{PQ}}} = \sqrt{\frac{80}{1/80}} = \sqrt{6400} = 80 \ m/s$.
The time taken to travel through wire $PQ$ is $t_1 = \frac{L_{PQ}}{v_{PQ}} = \frac{4.8}{80} = 0.06 \ s$.
The velocity of the wave in wire $QR$ is $v_{QR} = \sqrt{\frac{T}{m_{QR}}} = \sqrt{\frac{80}{5/64}} = \sqrt{\frac{80 \times 64}{5}} = \sqrt{16 \times 64} = 4 \times 8 = 32 \ m/s$.
The time taken to travel through wire $QR$ is $t_2 = \frac{L_{QR}}{v_{QR}} = \frac{2.56}{32} = 0.08 \ s$.
The total time taken is $t = t_1 + t_2 = 0.06 + 0.08 = 0.14 \ s$.
Mass per unit length for wire $QR$ is $m_{QR} = \frac{0.2}{2.56} = \frac{5}{64} \ kg/m$.
The velocity of the wave in wire $PQ$ is $v_{PQ} = \sqrt{\frac{T}{m_{PQ}}} = \sqrt{\frac{80}{1/80}} = \sqrt{6400} = 80 \ m/s$.
The time taken to travel through wire $PQ$ is $t_1 = \frac{L_{PQ}}{v_{PQ}} = \frac{4.8}{80} = 0.06 \ s$.
The velocity of the wave in wire $QR$ is $v_{QR} = \sqrt{\frac{T}{m_{QR}}} = \sqrt{\frac{80}{5/64}} = \sqrt{\frac{80 \times 64}{5}} = \sqrt{16 \times 64} = 4 \times 8 = 32 \ m/s$.
The time taken to travel through wire $QR$ is $t_2 = \frac{L_{QR}}{v_{QR}} = \frac{2.56}{32} = 0.08 \ s$.
The total time taken is $t = t_1 + t_2 = 0.06 + 0.08 = 0.14 \ s$.
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View Solution69
EasyMCQ
The equation of a simple harmonic wave produced in a string under a tension of $0.4 \,N$ is given by $y=4 \sin (3x+60t) \,m$. The mass per unit length of the string is:
A
$10^{-3} \,kg \,m^{-1}$
B
$10^{-5} \,kg \,m^{-1}$
C
$10^{-3} \,g \,cm^{-1}$
D
$10^{-5} \,g \,cm^{-1}$
Solution
(A) The standard equation of a traveling wave is given by $y=A \sin (kx+\omega t)$.
Comparing this with the given equation $y=4 \sin (3x+60t)$,we get the wave number $k=3 \,m^{-1}$ and the angular frequency $\omega=60 \,rad/s$.
The wave speed $V$ is calculated as $V = \frac{\omega}{k} = \frac{60}{3} = 20 \,m/s$.
The speed of a transverse wave in a stretched string is given by $V = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the mass per unit length.
Squaring both sides,we get $V^2 = \frac{T}{\mu}$,which implies $\mu = \frac{T}{V^2}$.
Substituting the given values $T = 0.4 \,N$ and $V = 20 \,m/s$:
$\mu = \frac{0.4}{(20)^2} = \frac{0.4}{400} = \frac{4 \times 10^{-1}}{4 \times 10^2} = 10^{-3} \,kg \,m^{-1}$.
Comparing this with the given equation $y=4 \sin (3x+60t)$,we get the wave number $k=3 \,m^{-1}$ and the angular frequency $\omega=60 \,rad/s$.
The wave speed $V$ is calculated as $V = \frac{\omega}{k} = \frac{60}{3} = 20 \,m/s$.
The speed of a transverse wave in a stretched string is given by $V = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the mass per unit length.
Squaring both sides,we get $V^2 = \frac{T}{\mu}$,which implies $\mu = \frac{T}{V^2}$.
Substituting the given values $T = 0.4 \,N$ and $V = 20 \,m/s$:
$\mu = \frac{0.4}{(20)^2} = \frac{0.4}{400} = \frac{4 \times 10^{-1}}{4 \times 10^2} = 10^{-3} \,kg \,m^{-1}$.
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View Solution70
DifficultMCQ
$A$ uniform rope of length $12 \ m$ and mass $6 \ kg$ hangs vertically from a rigid support. $A$ block of mass $2 \ kg$ is attached to the free end of the rope. $A$ transverse pulse of wavelength $0.06 \ m$ is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is (in $m$)
A
$0.12$
B
$0.4$
C
$0.8$
D
$0.16$
Solution
(A) The speed of a transverse wave in a string is given by $v = f \lambda = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Since the frequency $f$ of the pulse remains constant as it travels,we have $\lambda \propto \sqrt{T}$.
Therefore,the ratio of wavelengths is $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_2}{T_1}}$.
At the lower end (bottom),the tension $T_1$ is due to the block of mass $2 \ kg$: $T_1 = 2g$.
At the top of the rope,the tension $T_2$ supports both the block $(2 \ kg)$ and the rope $(6 \ kg)$: $T_2 = (2 + 6)g = 8g$.
Given $\lambda_1 = 0.06 \ m$,we calculate $\lambda_2$ as:
$\lambda_2 = \lambda_1 \sqrt{\frac{T_2}{T_1}} = 0.06 \times \sqrt{\frac{8g}{2g}} = 0.06 \times \sqrt{4} = 0.06 \times 2 = 0.12 \ m$.
Since the frequency $f$ of the pulse remains constant as it travels,we have $\lambda \propto \sqrt{T}$.
Therefore,the ratio of wavelengths is $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_2}{T_1}}$.
At the lower end (bottom),the tension $T_1$ is due to the block of mass $2 \ kg$: $T_1 = 2g$.
At the top of the rope,the tension $T_2$ supports both the block $(2 \ kg)$ and the rope $(6 \ kg)$: $T_2 = (2 + 6)g = 8g$.
Given $\lambda_1 = 0.06 \ m$,we calculate $\lambda_2$ as:
$\lambda_2 = \lambda_1 \sqrt{\frac{T_2}{T_1}} = 0.06 \times \sqrt{\frac{8g}{2g}} = 0.06 \times \sqrt{4} = 0.06 \times 2 = 0.12 \ m$.
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View Solution71
MediumMCQ
$A$ string of mass $0.1 \, kg$ is under a tension $1.6 \, N$. The length of the string is $1 \, m$. $A$ transverse wave starts from one end of the string. The time taken by the wave to reach the other end is (in $s$.)
A
$0.30$
B
$0.50$
C
$0.25$
D
$0.75$
Solution
(C) Given: Mass of string $m = 0.1 \, kg$, Tension $T = 1.6 \, N$, Length $\ell = 1 \, m$.
Mass per unit length $\mu = \frac{m}{\ell} = \frac{0.1}{1} = 0.1 \, kg/m$.
The velocity $v$ of the transverse wave in the string is given by the formula $v = \sqrt{\frac{T}{\mu}}$.
Substituting the values, $v = \sqrt{\frac{1.6}{0.1}} = \sqrt{16} = 4 \, m/s$.
The time taken $t$ to reach the other end is $t = \frac{\ell}{v} = \frac{1}{4} = 0.25 \, s$.
Mass per unit length $\mu = \frac{m}{\ell} = \frac{0.1}{1} = 0.1 \, kg/m$.
The velocity $v$ of the transverse wave in the string is given by the formula $v = \sqrt{\frac{T}{\mu}}$.
Substituting the values, $v = \sqrt{\frac{1.6}{0.1}} = \sqrt{16} = 4 \, m/s$.
The time taken $t$ to reach the other end is $t = \frac{\ell}{v} = \frac{1}{4} = 0.25 \, s$.
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View Solution72
MediumMCQ
$A$ transverse wave is travelling on a string with velocity $V$. The extension in the string is $x$. If the string is extended by $50 \%$,the speed of the wave along the string will be nearly (Hooke's law is obeyed).
A
$(0.7) V$
B
$(1.22) V$
C
$(1.1) V$
D
$(0.9) V$
Solution
(B) The speed of a transverse wave on a stretched string is given by $V = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the mass per unit length.
According to Hooke's Law,$T = kx$,where $k$ is the spring constant and $x$ is the extension.
Since the mass of the string remains constant,$\mu = \frac{m}{L}$. When the string is extended,its length $L$ increases. However,for small extensions,the change in $\mu$ is negligible compared to the change in tension $T$.
Thus,$V \propto \sqrt{T} \propto \sqrt{x}$.
Given the initial extension is $x_1 = x$ and the final extension is $x_2 = x + 0.5x = 1.5x$.
The ratio of the speeds is $\frac{V_2}{V_1} = \sqrt{\frac{x_2}{x_1}} = \sqrt{\frac{1.5x}{x}} = \sqrt{1.5}$.
Calculating the value,$\sqrt{1.5} \approx 1.2247$.
Therefore,the new speed $V_2 \approx 1.22 V$.
According to Hooke's Law,$T = kx$,where $k$ is the spring constant and $x$ is the extension.
Since the mass of the string remains constant,$\mu = \frac{m}{L}$. When the string is extended,its length $L$ increases. However,for small extensions,the change in $\mu$ is negligible compared to the change in tension $T$.
Thus,$V \propto \sqrt{T} \propto \sqrt{x}$.
Given the initial extension is $x_1 = x$ and the final extension is $x_2 = x + 0.5x = 1.5x$.
The ratio of the speeds is $\frac{V_2}{V_1} = \sqrt{\frac{x_2}{x_1}} = \sqrt{\frac{1.5x}{x}} = \sqrt{1.5}$.
Calculating the value,$\sqrt{1.5} \approx 1.2247$.
Therefore,the new speed $V_2 \approx 1.22 V$.
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View Solution73
MediumMCQ
$A$ uniform metal wire of length $L$,mass $M$,and density $\rho$ is under a tension $T$. If the speed of a transverse wave along the wire is $V$,then the area of cross-section of the wire is:
A
$\frac{V}{T\rho}$
B
$\frac{T}{V^{2}\rho}$
C
$\frac{T^{2}}{V\rho}$
D
$\frac{V^{2}}{T\rho}$
Solution
(B) The speed of a transverse wave in a stretched wire is given by $V = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Linear mass density $\mu = \frac{M}{L}$.
Also,mass $M = \text{Volume} \times \text{Density} = (A \times L) \times \rho$,where $A$ is the area of cross-section.
Therefore,$\mu = \frac{A \times L \times \rho}{L} = A\rho$.
Substituting this into the wave speed formula: $V = \sqrt{\frac{T}{A\rho}}$.
Squaring both sides: $V^{2} = \frac{T}{A\rho}$.
Rearranging for the area of cross-section $A$: $A = \frac{T}{V^{2}\rho}$.
Linear mass density $\mu = \frac{M}{L}$.
Also,mass $M = \text{Volume} \times \text{Density} = (A \times L) \times \rho$,where $A$ is the area of cross-section.
Therefore,$\mu = \frac{A \times L \times \rho}{L} = A\rho$.
Substituting this into the wave speed formula: $V = \sqrt{\frac{T}{A\rho}}$.
Squaring both sides: $V^{2} = \frac{T}{A\rho}$.
Rearranging for the area of cross-section $A$: $A = \frac{T}{V^{2}\rho}$.
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View Solution74
MediumMCQ
$A$ uniform metal wire has length $L$,mass $M$,and density $\rho$. It is under tension $T$,and $v$ is the speed of a transverse wave along the wire. The area of cross-section of the wire is:
A
$\frac{v^{2} \rho}{T}$
B
$\frac{T}{v^{2} \rho}$
C
$T^{2} \rho v$
D
$Tv^{2} \rho$
Solution
(B) The speed of a transverse wave on a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Linear mass density $\mu = \frac{M}{L}$.
Since mass $M = \text{Volume} \times \text{density} = (A \times L) \times \rho$,where $A$ is the cross-sectional area.
Therefore,$\mu = \frac{A \times L \times \rho}{L} = A \rho$.
Substituting this into the wave speed formula: $v = \sqrt{\frac{T}{A \rho}}$.
Squaring both sides: $v^{2} = \frac{T}{A \rho}$.
Rearranging for the area $A$: $A = \frac{T}{v^{2} \rho}$.
Linear mass density $\mu = \frac{M}{L}$.
Since mass $M = \text{Volume} \times \text{density} = (A \times L) \times \rho$,where $A$ is the cross-sectional area.
Therefore,$\mu = \frac{A \times L \times \rho}{L} = A \rho$.
Substituting this into the wave speed formula: $v = \sqrt{\frac{T}{A \rho}}$.
Squaring both sides: $v^{2} = \frac{T}{A \rho}$.
Rearranging for the area $A$: $A = \frac{T}{v^{2} \rho}$.
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View Solution75
EasyMCQ
Two $Cu$ wires of radii $R_{1}$ and $R_{2}$ are such that $(R_{1} > R_{2})$. Then which of the following is true?
A
Transverse wave travels faster in thicker wire
B
Transverse wave travels faster in thinner wire
C
Travels with the same speed in both the wires
D
Does not travel
Solution
(B) The velocity of a transverse wave in a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Since $\mu = \rho A = \rho (\pi R^2)$,where $\rho$ is the density of the material and $A$ is the cross-sectional area,we have:
$v = \sqrt{\frac{T}{\rho \pi R^2}} = \frac{1}{R} \sqrt{\frac{T}{\rho \pi}}$.
From this expression,it is clear that $v \propto \frac{1}{R}$.
Since $R_{1} > R_{2}$,the velocity $v_{1}$ in the thicker wire will be less than the velocity $v_{2}$ in the thinner wire $(v_{1} < v_{2})$.
Therefore,the transverse wave travels faster in the thinner wire.
Since $\mu = \rho A = \rho (\pi R^2)$,where $\rho$ is the density of the material and $A$ is the cross-sectional area,we have:
$v = \sqrt{\frac{T}{\rho \pi R^2}} = \frac{1}{R} \sqrt{\frac{T}{\rho \pi}}$.
From this expression,it is clear that $v \propto \frac{1}{R}$.
Since $R_{1} > R_{2}$,the velocity $v_{1}$ in the thicker wire will be less than the velocity $v_{2}$ in the thinner wire $(v_{1} < v_{2})$.
Therefore,the transverse wave travels faster in the thinner wire.
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View Solution76
MediumMCQ
$A$ uniform wire $20 \,m$ long and weighing $50 \,N$ hangs vertically. The speed of the wave at the midpoint of the wire is (acceleration due to gravity $= g = 10 \,ms^{-2}$)
A
$4 \,ms^{-1}$
B
$10 \sqrt{2} \,ms^{-1}$
C
$10 \,ms^{-1}$
D
Zero $ms^{-1}$
Solution
(C) The mass of the wire is $m = \frac{W}{g} = \frac{50}{10} = 5 \,kg$.
The linear mass density of the wire is $\mu = \frac{m}{L} = \frac{5}{20} = 0.25 \,kg/m$.
At the midpoint of the wire,the tension $T$ is equal to the weight of the lower half of the wire.
The mass of the lower half is $m' = \frac{m}{2} = 2.5 \,kg$.
Thus,$T = m'g = 2.5 \times 10 = 25 \,N$.
The speed of the wave $v$ is given by $v = \sqrt{\frac{T}{\mu}}$.
Substituting the values: $v = \sqrt{\frac{25}{0.25}} = \sqrt{100} = 10 \,ms^{-1}$.
The linear mass density of the wire is $\mu = \frac{m}{L} = \frac{5}{20} = 0.25 \,kg/m$.
At the midpoint of the wire,the tension $T$ is equal to the weight of the lower half of the wire.
The mass of the lower half is $m' = \frac{m}{2} = 2.5 \,kg$.
Thus,$T = m'g = 2.5 \times 10 = 25 \,N$.
The speed of the wave $v$ is given by $v = \sqrt{\frac{T}{\mu}}$.
Substituting the values: $v = \sqrt{\frac{25}{0.25}} = \sqrt{100} = 10 \,ms^{-1}$.
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View Solution77
MediumMCQ
$A$ string of mass $0.2 \ kg$ is under a tension of $2.5 \ N$. The length of the string is $2 \ m$. $A$ transverse wave starts from one end of the string. The time taken by the wave to reach the other end is: (in $s$)
A
$0.2$
B
$0.4$
C
$0.6$
D
$0.8$
Solution
(B) Mass per unit length $(m)$:
$m = \frac{M}{l} = \frac{0.2 \ kg}{2 \ m} = 0.1 \ kg/m$
Velocity of the transverse wave $(v)$:
$v = \sqrt{\frac{T}{m}} = \sqrt{\frac{2.5 \ N}{0.1 \ kg/m}} = \sqrt{25} = 5 \ m/s$
Time taken $(t)$ by the wave to reach the other end:
$t = \frac{l}{v} = \frac{2 \ m}{5 \ m/s} = 0.4 \ s$
$m = \frac{M}{l} = \frac{0.2 \ kg}{2 \ m} = 0.1 \ kg/m$
Velocity of the transverse wave $(v)$:
$v = \sqrt{\frac{T}{m}} = \sqrt{\frac{2.5 \ N}{0.1 \ kg/m}} = \sqrt{25} = 5 \ m/s$
Time taken $(t)$ by the wave to reach the other end:
$t = \frac{l}{v} = \frac{2 \ m}{5 \ m/s} = 0.4 \ s$
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View Solution78
MediumMCQ
$A$ uniform rope of length $L$ and mass $m_1$ hangs vertically from a rigid support. $A$ block of mass $m_2$ is attached to the free end of the rope. $A$ transverse wave of wavelength $\lambda_1$ is produced at the lower end of the rope. The wavelength of the wave when it reaches the top of the rope is $\lambda_2$. The ratio $\frac{\lambda_1}{\lambda_2}$ is
A
$\left[\frac{m_2}{m_1+m_2}\right]^{\frac{1}{2}}$
B
$\left[\frac{m_1+m_2}{m_2}\right]^{\frac{1}{2}}$
C
$\left[\frac{m_1}{m_1+m_2}\right]^{\frac{1}{2}}$
D
$\left[\frac{m_2}{m_1-m_2}\right]^{\frac{1}{2}}$
Solution
(A) Let the velocity of the pulse at the lower end be $v_1$ and at the top be $v_2$.
Since the frequency $n$ of the wave remains constant,we have $\lambda = \frac{v}{n}$,which implies $\frac{\lambda_2}{\lambda_1} = \frac{v_2}{v_1}$.
The velocity of a transverse wave on a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Thus,$v \propto \sqrt{T}$.
At the lower end,the tension $T_1$ is due to the block of mass $m_2$,so $T_1 = m_2 g$.
At the top end,the tension $T_2$ is due to the total mass of the rope and the block,so $T_2 = (m_1 + m_2) g$.
Therefore,$\frac{\lambda_2}{\lambda_1} = \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{(m_1 + m_2)g}{m_2 g}} = \sqrt{\frac{m_1 + m_2}{m_2}}$.
Taking the reciprocal,we get $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2}{m_1 + m_2}} = \left[\frac{m_2}{m_1+m_2}\right]^{\frac{1}{2}}$.
Since the frequency $n$ of the wave remains constant,we have $\lambda = \frac{v}{n}$,which implies $\frac{\lambda_2}{\lambda_1} = \frac{v_2}{v_1}$.
The velocity of a transverse wave on a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Thus,$v \propto \sqrt{T}$.
At the lower end,the tension $T_1$ is due to the block of mass $m_2$,so $T_1 = m_2 g$.
At the top end,the tension $T_2$ is due to the total mass of the rope and the block,so $T_2 = (m_1 + m_2) g$.
Therefore,$\frac{\lambda_2}{\lambda_1} = \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{(m_1 + m_2)g}{m_2 g}} = \sqrt{\frac{m_1 + m_2}{m_2}}$.
Taking the reciprocal,we get $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2}{m_1 + m_2}} = \left[\frac{m_2}{m_1+m_2}\right]^{\frac{1}{2}}$.
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View Solution79
EasyMCQ
Two copper wires of radii $r_1$ and $r_2$ $(r_1 > r_2)$ are subjected to the same tension and are plucked. The transverse waves will
A
not travel through both the wires
B
travel with the same velocity in both the wires
C
travel faster in the thinner wire
D
travel faster in the thicker wire
Solution
(C) The velocity $v$ of a transverse wave on a stretched string is given by the formula $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Linear mass density $\mu = \rho \cdot A$,where $\rho$ is the density of the material and $A$ is the cross-sectional area.
Since the wires are made of the same material (copper),$\rho$ is constant.
The cross-sectional area $A = \pi r^2$,so $\mu = \rho \cdot \pi r^2$.
Substituting this into the velocity formula: $v = \sqrt{\frac{T}{\rho \pi r^2}} = \frac{1}{r} \sqrt{\frac{T}{\rho \pi}}$.
Since $T$,$\rho$,and $\pi$ are constant,we have $v \propto \frac{1}{r}$.
Given $r_1 > r_2$,the thinner wire $(r_2)$ will have a higher velocity compared to the thicker wire $(r_1)$.
Therefore,the transverse wave travels faster in the thinner wire.
Linear mass density $\mu = \rho \cdot A$,where $\rho$ is the density of the material and $A$ is the cross-sectional area.
Since the wires are made of the same material (copper),$\rho$ is constant.
The cross-sectional area $A = \pi r^2$,so $\mu = \rho \cdot \pi r^2$.
Substituting this into the velocity formula: $v = \sqrt{\frac{T}{\rho \pi r^2}} = \frac{1}{r} \sqrt{\frac{T}{\rho \pi}}$.
Since $T$,$\rho$,and $\pi$ are constant,we have $v \propto \frac{1}{r}$.
Given $r_1 > r_2$,the thinner wire $(r_2)$ will have a higher velocity compared to the thicker wire $(r_1)$.
Therefore,the transverse wave travels faster in the thinner wire.
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View Solution80
MediumMCQ
$A$ uniform metal wire has length $L$,mass $M$,and cross-sectional area $A$. It is under tension $T$,and $V$ is the speed of a transverse wave along the wire. The density of the wire is:
A
$\frac{A T}{V^2}$
B
$\frac{T}{A^2 V}$
C
$\frac{T}{V^2 A}$
D
$\frac{V^2}{A^2 T}$
Solution
(C) The speed of a transverse wave on a stretched wire is given by $V = \sqrt{\frac{T}{\mu}}$,where $\mu$ is the mass per unit length.
Given that the wire has mass $M$,length $L$,and cross-sectional area $A$,the mass per unit length $\mu$ is $\mu = \frac{M}{L}$.
Since $M = \text{density} (\rho) \times \text{volume} = \rho \times A \times L$,we have $\mu = \frac{\rho A L}{L} = \rho A$.
Substituting this into the wave speed formula: $V = \sqrt{\frac{T}{\rho A}}$.
Squaring both sides: $V^2 = \frac{T}{\rho A}$.
Rearranging to solve for density $\rho$: $\rho = \frac{T}{V^2 A}$.
Given that the wire has mass $M$,length $L$,and cross-sectional area $A$,the mass per unit length $\mu$ is $\mu = \frac{M}{L}$.
Since $M = \text{density} (\rho) \times \text{volume} = \rho \times A \times L$,we have $\mu = \frac{\rho A L}{L} = \rho A$.
Substituting this into the wave speed formula: $V = \sqrt{\frac{T}{\rho A}}$.
Squaring both sides: $V^2 = \frac{T}{\rho A}$.
Rearranging to solve for density $\rho$: $\rho = \frac{T}{V^2 A}$.
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View Solution81
MediumMCQ
$A$ string of mass $M$ is under a tension $T$. The length of the string is $L$. $A$ transverse wave starts at one end of the string. The time required for the disturbance to reach the other end is
A
$\sqrt{\frac{L M}{T}}$
B
$\sqrt{L M T}$
C
$\sqrt{\frac{T}{L M}}$
D
$\sqrt{\frac{L T}{M}}$
Solution
(A) The velocity of a transverse wave on a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the mass per unit length of the string,defined as $\mu = \frac{M}{L}$.
Substituting $\mu$ into the velocity formula,we get $v = \sqrt{\frac{T}{M/L}} = \sqrt{\frac{TL}{M}}$.
The time $t$ required for the disturbance to travel the length $L$ of the string is given by $t = \frac{L}{v}$.
Substituting the expression for $v$,we get $t = \frac{L}{\sqrt{TL/M}} = L \sqrt{\frac{M}{TL}} = \sqrt{\frac{L^2 M}{TL}} = \sqrt{\frac{LM}{T}}$.
Substituting $\mu$ into the velocity formula,we get $v = \sqrt{\frac{T}{M/L}} = \sqrt{\frac{TL}{M}}$.
The time $t$ required for the disturbance to travel the length $L$ of the string is given by $t = \frac{L}{v}$.
Substituting the expression for $v$,we get $t = \frac{L}{\sqrt{TL/M}} = L \sqrt{\frac{M}{TL}} = \sqrt{\frac{L^2 M}{TL}} = \sqrt{\frac{LM}{T}}$.
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View Solution82
MediumMCQ
$A$ transverse wave is propagating on a string. The linear mass density of the vibrating string is $10^{-3} \ kg/m$. The equation of the wave is $Y = 0.05 \sin(x + 15t)$,where $x$ and $Y$ are in meters and time $t$ is in seconds. The tension in the string is: (in $N$)
A
$0.2$
B
$0.250$
C
$0.225$
D
$0.325$
Solution
(C) The general equation of a transverse wave is given by $Y = A \sin(kx + \omega t)$.
Comparing this with the given equation $Y = 0.05 \sin(x + 15t)$,we get the angular frequency $\omega = 15 \ rad/s$ and the wave number $k = 1 \ m^{-1}$.
The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{15}{1} = 15 \ m/s$.
The speed of a transverse wave on a string is also given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Given $\mu = 10^{-3} \ kg/m$,we have $15 = \sqrt{\frac{T}{10^{-3}}}$.
Squaring both sides,$225 = \frac{T}{10^{-3}}$.
Therefore,$T = 225 \times 10^{-3} \ N = 0.225 \ N$.
Comparing this with the given equation $Y = 0.05 \sin(x + 15t)$,we get the angular frequency $\omega = 15 \ rad/s$ and the wave number $k = 1 \ m^{-1}$.
The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{15}{1} = 15 \ m/s$.
The speed of a transverse wave on a string is also given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Given $\mu = 10^{-3} \ kg/m$,we have $15 = \sqrt{\frac{T}{10^{-3}}}$.
Squaring both sides,$225 = \frac{T}{10^{-3}}$.
Therefore,$T = 225 \times 10^{-3} \ N = 0.225 \ N$.
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View Solution83
EasyMCQ
Two strings $A$ and $B$ of the same material are stretched by the same tension. The radius of string $A$ is double the radius of string $B$. $A$ transverse wave travels on string $A$ with speed $V_A$ and on string $B$ with speed $V_B$. The ratio $\frac{V_A}{V_B}$ is:
A
$\frac{1}{4}$
B
$\frac{1}{2}$
C
$2$
D
$4$
Solution
(B) The velocity $V$ of a transverse wave travelling on a stretched string is given by the formula $V = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Since $\mu = \rho \cdot A = \rho \cdot \pi r^2$,where $\rho$ is the density of the material and $r$ is the radius of the string,we have $V = \sqrt{\frac{T}{\rho \cdot \pi r^2}}$.
Given that the material is the same ($\rho$ is constant) and the tension $T$ is the same,we find that $V \propto \frac{1}{r}$.
Therefore,the ratio of the speeds is $\frac{V_A}{V_B} = \frac{r_B}{r_A}$.
Given $r_A = 2r_B$,we substitute this into the ratio: $\frac{V_A}{V_B} = \frac{r_B}{2r_B} = \frac{1}{2}$.
Since $\mu = \rho \cdot A = \rho \cdot \pi r^2$,where $\rho$ is the density of the material and $r$ is the radius of the string,we have $V = \sqrt{\frac{T}{\rho \cdot \pi r^2}}$.
Given that the material is the same ($\rho$ is constant) and the tension $T$ is the same,we find that $V \propto \frac{1}{r}$.
Therefore,the ratio of the speeds is $\frac{V_A}{V_B} = \frac{r_B}{r_A}$.
Given $r_A = 2r_B$,we substitute this into the ratio: $\frac{V_A}{V_B} = \frac{r_B}{2r_B} = \frac{1}{2}$.
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View Solution84
EasyMCQ
$A$ metallic wire of $1 \ m$ length has a mass of $10 \times 10^{-3} \ kg$. If a tension of $100 \ N$ is applied to the wire,what is the speed of the transverse wave (in $ms^{-1}$)?
A
$100$
B
$10$
C
$200$
D
$0.1$
Solution
(A) Given: Length of wire,$l = 1 \ m$; Mass of wire,$m = 10 \times 10^{-3} \ kg$; Tension,$T = 100 \ N$.
The speed of a transverse wave in a stretched string is given by the formula: $v = \sqrt{\frac{T}{\mu}}$,where $\mu$ is the linear mass density.
Linear mass density $\mu = \frac{m}{l} = \frac{10 \times 10^{-3} \ kg}{1 \ m} = 10 \times 10^{-3} \ kg/m$.
Substituting the values into the formula:
$v = \sqrt{\frac{100}{10 \times 10^{-3}}} = \sqrt{\frac{100}{0.01}} = \sqrt{10000} = 100 \ ms^{-1}$.
Therefore,the speed of the transverse wave is $100 \ ms^{-1}$.
The speed of a transverse wave in a stretched string is given by the formula: $v = \sqrt{\frac{T}{\mu}}$,where $\mu$ is the linear mass density.
Linear mass density $\mu = \frac{m}{l} = \frac{10 \times 10^{-3} \ kg}{1 \ m} = 10 \times 10^{-3} \ kg/m$.
Substituting the values into the formula:
$v = \sqrt{\frac{100}{10 \times 10^{-3}}} = \sqrt{\frac{100}{0.01}} = \sqrt{10000} = 100 \ ms^{-1}$.
Therefore,the speed of the transverse wave is $100 \ ms^{-1}$.
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View Solution85
MediumMCQ
$A$ steel wire of length $81 \text{ cm}$ has a mass of $5 \times 10^{-3} \text{ kg}$. If the wire is under a tension of $50 \text{ N}$, then the speed of transverse waves on the wire is (in $\text{ m s}^{-1}$)
A
$100$
B
$105$
C
$90$
D
$60$
Solution
(C) The speed of a transverse wave on a stretched string is given by the formula $v = \sqrt{\frac{T}{\mu}}$, where $T$ is the tension and $\mu$ is the linear mass density.
Linear mass density $\mu = \frac{m}{L}$.
Given: $m = 5 \times 10^{-3} \text{ kg}$, $L = 81 \text{ cm} = 0.81 \text{ m}$, $T = 50 \text{ N}$.
Calculate $\mu$: $\mu = \frac{5 \times 10^{-3}}{0.81} \text{ kg/m}$.
Now, substitute the values into the speed formula:
$v = \sqrt{\frac{50}{\frac{5 \times 10^{-3}}{0.81}}} = \sqrt{\frac{50 \times 0.81}{5 \times 10^{-3}}} = \sqrt{10 \times 0.81 \times 10^3} = \sqrt{8100} = 90 \text{ m s}^{-1}$.
Thus, the speed of the transverse wave is $90 \text{ m s}^{-1}$.
Linear mass density $\mu = \frac{m}{L}$.
Given: $m = 5 \times 10^{-3} \text{ kg}$, $L = 81 \text{ cm} = 0.81 \text{ m}$, $T = 50 \text{ N}$.
Calculate $\mu$: $\mu = \frac{5 \times 10^{-3}}{0.81} \text{ kg/m}$.
Now, substitute the values into the speed formula:
$v = \sqrt{\frac{50}{\frac{5 \times 10^{-3}}{0.81}}} = \sqrt{\frac{50 \times 0.81}{5 \times 10^{-3}}} = \sqrt{10 \times 0.81 \times 10^3} = \sqrt{8100} = 90 \text{ m s}^{-1}$.
Thus, the speed of the transverse wave is $90 \text{ m s}^{-1}$.
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View Solution86
MediumMCQ
The equation of a transverse wave propagating along a stretched string of length $80 \ cm$ is $y=1.5 \sin \{(5 \times 10^{-3} x) + 20 t\}$,where $x$ and $y$ are in $cm$ and the time $t$ is in seconds. If the mass of the string is $3 \ g$,then the tension in the string is: (in $N$)
A
$12$
B
$4$
C
$6$
D
$8$
Solution
(C) The standard equation of a wave is $y = A \sin(kx + \omega t)$.
Comparing this with the given equation $y = 1.5 \sin((5 \times 10^{-3} x) + 20 t)$,we get:
Wave number $k = 5 \times 10^{-3} \ cm^{-1}$
Angular frequency $\omega = 20 \ rad/s$
The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{20}{5 \times 10^{-3}} = 4000 \ cm/s = 40 \ m/s$.
The mass per unit length $\mu$ is $\frac{m}{L} = \frac{3 \ g}{80 \ cm} = \frac{3 \times 10^{-3} \ kg}{0.8 \ m} = 3.75 \times 10^{-3} \ kg/m$.
The wave speed in a stretched string is $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension.
Therefore,$T = v^2 \mu = (40)^2 \times (3.75 \times 10^{-3}) = 1600 \times 0.00375 = 6 \ N$.
Comparing this with the given equation $y = 1.5 \sin((5 \times 10^{-3} x) + 20 t)$,we get:
Wave number $k = 5 \times 10^{-3} \ cm^{-1}$
Angular frequency $\omega = 20 \ rad/s$
The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{20}{5 \times 10^{-3}} = 4000 \ cm/s = 40 \ m/s$.
The mass per unit length $\mu$ is $\frac{m}{L} = \frac{3 \ g}{80 \ cm} = \frac{3 \times 10^{-3} \ kg}{0.8 \ m} = 3.75 \times 10^{-3} \ kg/m$.
The wave speed in a stretched string is $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension.
Therefore,$T = v^2 \mu = (40)^2 \times (3.75 \times 10^{-3}) = 1600 \times 0.00375 = 6 \ N$.
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View Solution87
MediumMCQ
The speed of a wave on a string is $150 \,ms^{-1}$ when the tension is $120 \,N$. The percentage increase in the tension in order to raise the wave speed by $20 \%$ is
A
$44$
B
$40$
C
$20$
D
$22$
Solution
(A) For a wave on a string, the speed $v$ is given by $v = \sqrt{\frac{T}{\mu}}$, where $T$ is the tension and $\mu$ is the linear mass density.
Since $\mu$ is constant, $v \propto \sqrt{T}$.
Given $v_1 = 150 \,ms^{-1}$ and $T_1 = 120 \,N$.
We want to increase the speed by $20 \%$, so $v_2 = v_1 + 0.20 v_1 = 1.2 v_1$.
Using the proportionality $v \propto \sqrt{T}$, we have $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $1.2 = \sqrt{\frac{T_2}{120}}$.
Squaring both sides: $1.44 = \frac{T_2}{120}$.
$T_2 = 1.44 \times 120 = 172.8 \,N$.
The percentage increase in tension is given by $\frac{T_2 - T_1}{T_1} \times 100$.
$\% \Delta T = \frac{172.8 - 120}{120} \times 100 = \frac{52.8}{120} \times 100 = 44 \%$.
Since $\mu$ is constant, $v \propto \sqrt{T}$.
Given $v_1 = 150 \,ms^{-1}$ and $T_1 = 120 \,N$.
We want to increase the speed by $20 \%$, so $v_2 = v_1 + 0.20 v_1 = 1.2 v_1$.
Using the proportionality $v \propto \sqrt{T}$, we have $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $1.2 = \sqrt{\frac{T_2}{120}}$.
Squaring both sides: $1.44 = \frac{T_2}{120}$.
$T_2 = 1.44 \times 120 = 172.8 \,N$.
The percentage increase in tension is given by $\frac{T_2 - T_1}{T_1} \times 100$.
$\% \Delta T = \frac{172.8 - 120}{120} \times 100 = \frac{52.8}{120} \times 100 = 44 \%$.
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View Solution88
MediumMCQ
$A$ string of length $L$ is stretched by $\frac{L}{20}$ and the speed of transverse waves along it is $v$. The speed of the wave when it is stretched by $\frac{L}{10}$ will be (assume that Hooke's law is applicable).
A
$2 v$
B
$\frac{v}{\sqrt{2}}$
C
$v \sqrt{2}$
D
$4 v$
Solution
(C) The speed of a transverse wave in a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
According to Hooke's Law,the tension $T$ in a stretched string is proportional to the extension $\Delta l$,so $T = k \Delta l$.
Since the mass and length of the string remain constant,$\mu$ is constant. Thus,$v \propto \sqrt{T} \propto \sqrt{\Delta l}$.
Given the initial extension $\Delta l_1 = \frac{L}{20}$ and initial speed $v_1 = v$.
For the second case,the extension is $\Delta l_2 = \frac{L}{10}$.
Using the ratio: $\frac{v_2}{v_1} = \sqrt{\frac{\Delta l_2}{\Delta l_1}} = \sqrt{\frac{L/10}{L/20}} = \sqrt{\frac{20}{10}} = \sqrt{2}$.
Therefore,$v_2 = v \sqrt{2}$.
According to Hooke's Law,the tension $T$ in a stretched string is proportional to the extension $\Delta l$,so $T = k \Delta l$.
Since the mass and length of the string remain constant,$\mu$ is constant. Thus,$v \propto \sqrt{T} \propto \sqrt{\Delta l}$.
Given the initial extension $\Delta l_1 = \frac{L}{20}$ and initial speed $v_1 = v$.
For the second case,the extension is $\Delta l_2 = \frac{L}{10}$.
Using the ratio: $\frac{v_2}{v_1} = \sqrt{\frac{\Delta l_2}{\Delta l_1}} = \sqrt{\frac{L/10}{L/20}} = \sqrt{\frac{20}{10}} = \sqrt{2}$.
Therefore,$v_2 = v \sqrt{2}$.
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View Solution89
MediumMCQ
The ratio of radii of two wires is $1: 2$ and the ratio of the densities of their materials is $1: 4$. If the same tension is applied to both wires,then the ratio of the speed of transverse waves produced in them is
A
$1: 16$
B
$16: 1$
C
$1: 4$
D
$4: 1$
Solution
(D) The speed of a transverse wave in a stretched wire is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Linear mass density $\mu = \frac{\text{mass}}{\text{length}} = \frac{\rho \times \text{Volume}}{\text{length}} = \rho \times A = \rho \times \pi R^2$,where $\rho$ is the material density and $R$ is the radius of the wire.
Substituting $\mu$ in the velocity formula: $v = \sqrt{\frac{T}{\rho \pi R^2}}$.
Given the ratio of radii $\frac{R_1}{R_2} = \frac{1}{2}$ and the ratio of densities $\frac{\rho_1}{\rho_2} = \frac{1}{4}$.
Since the tension $T$ is the same for both wires,the ratio of speeds is:
$\frac{v_1}{v_2} = \sqrt{\frac{\rho_2}{\rho_1} \times \frac{R_2^2}{R_1^2}}$
$\frac{v_1}{v_2} = \sqrt{\left(\frac{4}{1}\right) \times \left(\frac{2}{1}\right)^2} = \sqrt{4 \times 4} = \sqrt{16} = 4$.
Thus,the ratio is $4: 1$.
Linear mass density $\mu = \frac{\text{mass}}{\text{length}} = \frac{\rho \times \text{Volume}}{\text{length}} = \rho \times A = \rho \times \pi R^2$,where $\rho$ is the material density and $R$ is the radius of the wire.
Substituting $\mu$ in the velocity formula: $v = \sqrt{\frac{T}{\rho \pi R^2}}$.
Given the ratio of radii $\frac{R_1}{R_2} = \frac{1}{2}$ and the ratio of densities $\frac{\rho_1}{\rho_2} = \frac{1}{4}$.
Since the tension $T$ is the same for both wires,the ratio of speeds is:
$\frac{v_1}{v_2} = \sqrt{\frac{\rho_2}{\rho_1} \times \frac{R_2^2}{R_1^2}}$
$\frac{v_1}{v_2} = \sqrt{\left(\frac{4}{1}\right) \times \left(\frac{2}{1}\right)^2} = \sqrt{4 \times 4} = \sqrt{16} = 4$.
Thus,the ratio is $4: 1$.
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View Solution90
MediumMCQ
When the area of cross-section of a stretched wire is halved and tension is doubled,the speed of propagation of transverse waves along it becomes $k$ times the initial speed. Then,$k$ is:
A
$1$
B
$4$
C
$2$
D
$8$
Solution
(C) The speed of a transverse wave in a stretched wire is given by the formula:
$v = \sqrt{\frac{T}{\mu}}$
where $T$ is the tension in the wire and $\mu$ is the mass per unit length.
Since $\mu = \frac{M}{l} = \frac{A \cdot l \cdot \rho}{l} = A \cdot \rho$,where $A$ is the cross-sectional area and $\rho$ is the density of the material,the speed becomes:
$v = \sqrt{\frac{T}{A \rho}}$
This implies $v \propto \sqrt{\frac{T}{A}}$.
Given that the new area $A_2 = \frac{A_1}{2}$ and the new tension $T_2 = 2T_1$,the ratio of the new speed $v_2$ to the initial speed $v_1$ is:
$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1} \times \frac{A_1}{A_2}}$
Substituting the given values:
$\frac{v_2}{v_1} = \sqrt{\frac{2T_1}{T_1} \times \frac{A_1}{A_1/2}} = \sqrt{2 \times 2} = \sqrt{4} = 2$
Thus,$v_2 = 2v_1$,which means $k = 2$.
$v = \sqrt{\frac{T}{\mu}}$
where $T$ is the tension in the wire and $\mu$ is the mass per unit length.
Since $\mu = \frac{M}{l} = \frac{A \cdot l \cdot \rho}{l} = A \cdot \rho$,where $A$ is the cross-sectional area and $\rho$ is the density of the material,the speed becomes:
$v = \sqrt{\frac{T}{A \rho}}$
This implies $v \propto \sqrt{\frac{T}{A}}$.
Given that the new area $A_2 = \frac{A_1}{2}$ and the new tension $T_2 = 2T_1$,the ratio of the new speed $v_2$ to the initial speed $v_1$ is:
$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1} \times \frac{A_1}{A_2}}$
Substituting the given values:
$\frac{v_2}{v_1} = \sqrt{\frac{2T_1}{T_1} \times \frac{A_1}{A_1/2}} = \sqrt{2 \times 2} = \sqrt{4} = 2$
Thus,$v_2 = 2v_1$,which means $k = 2$.
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View Solution91
MediumMCQ
As shown in the figure, a block of mass $9 \, kg$ is hung by a wire of area of cross-section $1 \, mm^2$ in a lift going up with an acceleration of $2 \, ms^{-2}$. If the speed of the transverse wave on the wire is $120 \, ms^{-1}$, the density of the material of the wire is (Acceleration due to gravity $= 10 \, ms^{-2}$)
A
$1.5 \, g \, cm^{-3}$
B
$3.5 \, g \, cm^{-3}$
C
$5.5 \, g \, cm^{-3}$
D
$7.5 \, g \, cm^{-3}$
Solution
(D) The tension $T$ in the wire when the lift is accelerating upwards is given by $T = m(g + a)$.
Given $m = 9 \, kg$, $g = 10 \, ms^{-2}$, and $a = 2 \, ms^{-2}$, we have $T = 9(10 + 2) = 9 \times 12 = 108 \, N$.
The speed of a transverse wave on a wire is given by $v = \sqrt{\frac{T}{\mu}}$, where $\mu$ is the linear mass density.
We know $\mu = \rho A$, where $\rho$ is the density of the material and $A$ is the cross-sectional area.
Given $v = 120 \, ms^{-1}$ and $A = 1 \, mm^2 = 10^{-6} \, m^2$, we have $v^2 = \frac{T}{\rho A}$.
Rearranging for $\rho$, we get $\rho = \frac{T}{v^2 A} = \frac{108}{(120)^2 \times 10^{-6}} = \frac{108}{14400 \times 10^{-6}} = \frac{108}{0.0144} = 7500 \, kg \, m^{-3}$.
Converting to $g \, cm^{-3}$, $\rho = 7500 \times 10^{-3} \, g \, cm^{-3} = 7.5 \, g \, cm^{-3}$.
Thus, the correct option is $D$.
Given $m = 9 \, kg$, $g = 10 \, ms^{-2}$, and $a = 2 \, ms^{-2}$, we have $T = 9(10 + 2) = 9 \times 12 = 108 \, N$.
The speed of a transverse wave on a wire is given by $v = \sqrt{\frac{T}{\mu}}$, where $\mu$ is the linear mass density.
We know $\mu = \rho A$, where $\rho$ is the density of the material and $A$ is the cross-sectional area.
Given $v = 120 \, ms^{-1}$ and $A = 1 \, mm^2 = 10^{-6} \, m^2$, we have $v^2 = \frac{T}{\rho A}$.
Rearranging for $\rho$, we get $\rho = \frac{T}{v^2 A} = \frac{108}{(120)^2 \times 10^{-6}} = \frac{108}{14400 \times 10^{-6}} = \frac{108}{0.0144} = 7500 \, kg \, m^{-3}$.
Converting to $g \, cm^{-3}$, $\rho = 7500 \times 10^{-3} \, g \, cm^{-3} = 7.5 \, g \, cm^{-3}$.
Thus, the correct option is $D$.
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View Solution92
MediumMCQ
$A$ string of length $1 \,m$ and mass $490 \,g$ is put under a tension of $25 \,N$. $A$ wave of frequency $120 \,Hz$ is sent along it. The speed of this wave is (in $\,m/s$)
A
$7.14$
B
$0.71$
C
$0.51$
D
$51.0$
Solution
(A) The speed of a transverse wave on a stretched string is given by the formula $v = \sqrt{\frac{T}{\mu}}$, where $T$ is the tension and $\mu$ is the linear mass density.
Given:
Tension $T = 25 \,N$
Length $L = 1 \,m$
Mass $M = 490 \,g = 0.49 \,kg$
Linear mass density $\mu = \frac{M}{L} = \frac{0.49 \,kg}{1 \,m} = 0.49 \,kg/m$.
Substituting the values into the formula:
$v = \sqrt{\frac{25}{0.49}} = \sqrt{\frac{2500}{49}} = \frac{50}{7} \approx 7.14 \,m/s$.
Given:
Tension $T = 25 \,N$
Length $L = 1 \,m$
Mass $M = 490 \,g = 0.49 \,kg$
Linear mass density $\mu = \frac{M}{L} = \frac{0.49 \,kg}{1 \,m} = 0.49 \,kg/m$.
Substituting the values into the formula:
$v = \sqrt{\frac{25}{0.49}} = \sqrt{\frac{2500}{49}} = \frac{50}{7} \approx 7.14 \,m/s$.
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View Solution93
DifficultMCQ
$A$ heavy uniform rope is suspended vertically from a ceiling and is in equilibrium. $A$ pulse is generated at the bottom end of the rope as shown. As the pulse travels up the rope,its acceleration at any instant is ($g$ is acceleration due to gravity).
A
Constant and equal to $\frac{g}{2}$
B
Variable but equal to $\frac{g}{2}$ when the pulse is exactly at the middle of the string
C
Constant and equal to $g$
D
Variable but equal to $g$ when the pulse is exactly at the middle of the string.
Solution
(A) Let the total mass of the rope be $M$ and its length be $L$. Let $\mu = \frac{M}{L}$ be the mass per unit length.
Consider the pulse at a distance $x$ from the bottom end of the rope.
The tension $T$ at a distance $x$ from the bottom is equal to the weight of the rope segment of length $x$ below that point: $T = \mu x g$.
The speed of the pulse $v$ is given by $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{\mu x g}{\mu}} = \sqrt{xg}$.
The acceleration of the pulse is $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}$.
Since $v = \sqrt{xg}$,we have $\frac{dv}{dx} = \frac{1}{2\sqrt{x}} \sqrt{g}$.
Also,$\frac{dx}{dt} = v = \sqrt{xg}$.
Therefore,$a = \left( \frac{1}{2\sqrt{x}} \sqrt{g} \right) \cdot \sqrt{xg} = \frac{1}{2} \sqrt{g} \cdot \sqrt{g} = \frac{g}{2}$.
Since the acceleration $a = \frac{g}{2}$ is independent of $x$,the acceleration of the pulse is constant throughout its motion.
Consider the pulse at a distance $x$ from the bottom end of the rope.
The tension $T$ at a distance $x$ from the bottom is equal to the weight of the rope segment of length $x$ below that point: $T = \mu x g$.
The speed of the pulse $v$ is given by $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{\mu x g}{\mu}} = \sqrt{xg}$.
The acceleration of the pulse is $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}$.
Since $v = \sqrt{xg}$,we have $\frac{dv}{dx} = \frac{1}{2\sqrt{x}} \sqrt{g}$.
Also,$\frac{dx}{dt} = v = \sqrt{xg}$.
Therefore,$a = \left( \frac{1}{2\sqrt{x}} \sqrt{g} \right) \cdot \sqrt{xg} = \frac{1}{2} \sqrt{g} \cdot \sqrt{g} = \frac{g}{2}$.
Since the acceleration $a = \frac{g}{2}$ is independent of $x$,the acceleration of the pulse is constant throughout its motion.
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View Solution94
MediumMCQ
The equation of a transverse wave propagating on a stretched string is given by $y = 3 \sin (4x + 200t)$,where $x$ and $y$ are in metres and the time $t$ is in seconds. If the tension applied to the string is $500 \ N$,the linear density of the string is: (in $kg \ m^{-1}$)
A
$0.25$
B
$0.4$
C
$0.2$
D
$0.1$
Solution
(C) The standard equation of a transverse wave is $y = A \sin(kx + \omega t)$.
Comparing this with the given equation $y = 3 \sin(4x + 200t)$,we get:
Wave number $k = 4 \ m^{-1}$
Angular frequency $\omega = 200 \ rad \ s^{-1}$
The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{200}{4} = 50 \ m/s$.
The speed of a transverse wave on a stretched string is also given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Squaring both sides,we get $v^2 = \frac{T}{\mu}$,which implies $\mu = \frac{T}{v^2}$.
Substituting the given values $T = 500 \ N$ and $v = 50 \ m/s$:
$\mu = \frac{500}{(50)^2} = \frac{500}{2500} = \frac{1}{5} = 0.2 \ kg \ m^{-1}$.
Thus,the linear density of the string is $0.2 \ kg \ m^{-1}$.
Comparing this with the given equation $y = 3 \sin(4x + 200t)$,we get:
Wave number $k = 4 \ m^{-1}$
Angular frequency $\omega = 200 \ rad \ s^{-1}$
The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{200}{4} = 50 \ m/s$.
The speed of a transverse wave on a stretched string is also given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Squaring both sides,we get $v^2 = \frac{T}{\mu}$,which implies $\mu = \frac{T}{v^2}$.
Substituting the given values $T = 500 \ N$ and $v = 50 \ m/s$:
$\mu = \frac{500}{(50)^2} = \frac{500}{2500} = \frac{1}{5} = 0.2 \ kg \ m^{-1}$.
Thus,the linear density of the string is $0.2 \ kg \ m^{-1}$.
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View Solution95
DifficultMCQ
The speed of a transverse wave in a stretched string '$A$' is '$v$'. Another string '$B$' of the same length and same radius is subjected to the same tension. If the density of the material of the string '$B$' is $2\%$ more than that of '$A$',then the speed of the transverse wave in string '$B$' is
A
$\sqrt{1.04} v$
B
$\sqrt{1.02} v$
C
$\frac{v}{\sqrt{1.04}}$
D
$\frac{v}{\sqrt{1.02}}$
Solution
(D) The speed of a transverse wave in a stretched string is given by the formula $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Since $\mu = \text{Area} \times \text{density} = (\pi r^2) \rho$,the speed becomes $v = \sqrt{\frac{T}{\pi r^2 \rho}}$.
Given that the tension $T$ and radius $r$ are the same for both strings,we have $v \propto \frac{1}{\sqrt{\rho}}$.
Let $\rho_A = \rho$. Then $\rho_B = \rho + 0.02\rho = 1.02\rho$.
The speed in string '$B$' is $v_B = \sqrt{\frac{T}{\pi r^2 (1.02\rho)}} = \frac{1}{\sqrt{1.02}} \sqrt{\frac{T}{\pi r^2 \rho}}$.
Therefore,$v_B = \frac{v}{\sqrt{1.02}}$.
Since $\mu = \text{Area} \times \text{density} = (\pi r^2) \rho$,the speed becomes $v = \sqrt{\frac{T}{\pi r^2 \rho}}$.
Given that the tension $T$ and radius $r$ are the same for both strings,we have $v \propto \frac{1}{\sqrt{\rho}}$.
Let $\rho_A = \rho$. Then $\rho_B = \rho + 0.02\rho = 1.02\rho$.
The speed in string '$B$' is $v_B = \sqrt{\frac{T}{\pi r^2 (1.02\rho)}} = \frac{1}{\sqrt{1.02}} \sqrt{\frac{T}{\pi r^2 \rho}}$.
Therefore,$v_B = \frac{v}{\sqrt{1.02}}$.
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View Solution96
DifficultMCQ
$A$ uniform rope of mass $0.1 \,kg$ and length $2.45 \,m$ hangs from a rigid support. The time taken by a transverse wave formed in the rope to travel through the full length of the rope is (Assume $g = 9.8 \,m/s^2$). (in $\,s$)
A
$0.5$
B
$1.6$
C
$1.2$
D
$1.0$
Solution
(D) The velocity of a transverse wave in a rope at a distance $x$ from the free end is given by $v = \sqrt{gx}$.
Since $v = \frac{dx}{dt}$, we have $\frac{dx}{dt} = \sqrt{gx}$.
Rearranging the terms, we get $dt = \frac{dx}{\sqrt{gx}}$.
Integrating from $x = 0$ to $x = l$, the total time $t$ is:
$t = \int_{0}^{l} \frac{dx}{\sqrt{gx}} = \frac{1}{\sqrt{g}} [2\sqrt{x}]_{0}^{l} = 2\sqrt{\frac{l}{g}}$.
Substituting the given values $l = 2.45 \,m$ and $g = 9.8 \,m/s^2$:
$t = 2 \sqrt{\frac{2.45}{9.8}} = 2 \sqrt{\frac{1}{4}} = 2 \times 0.5 = 1 \,s$.
Since $v = \frac{dx}{dt}$, we have $\frac{dx}{dt} = \sqrt{gx}$.
Rearranging the terms, we get $dt = \frac{dx}{\sqrt{gx}}$.
Integrating from $x = 0$ to $x = l$, the total time $t$ is:
$t = \int_{0}^{l} \frac{dx}{\sqrt{gx}} = \frac{1}{\sqrt{g}} [2\sqrt{x}]_{0}^{l} = 2\sqrt{\frac{l}{g}}$.
Substituting the given values $l = 2.45 \,m$ and $g = 9.8 \,m/s^2$:
$t = 2 \sqrt{\frac{2.45}{9.8}} = 2 \sqrt{\frac{1}{4}} = 2 \times 0.5 = 1 \,s$.
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View Solution97
DifficultMCQ
$A$ transverse wave propagating on a stretched string of linear density $3 \times 10^{-4} \ kg \ m^{-1}$ is represented by the equation $y = 0.2 \sin (1.5 x + 60 t)$,where $x$ is in metres and $t$ is in seconds. The tension in the string (in newton) is
A
$0.24$
B
$0.48$
C
$1.2$
D
$1.8$
Solution
(B) The given equation of the wave is $y = 0.2 \sin (1.5 x + 60 t)$.
Comparing this with the standard wave equation $y = A \sin (kx + \omega t)$,we get:
Wave number $k = 1.5 \ m^{-1}$ and angular frequency $\omega = 60 \ rad \ s^{-1}$.
The velocity of the wave is given by $v = \frac{\omega}{k} = \frac{60}{1.5} = 40 \ m \ s^{-1}$.
The velocity of a transverse wave in a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Given $\mu = 3 \times 10^{-4} \ kg \ m^{-1}$.
Rearranging for tension $T$: $T = v^2 \mu$.
Substituting the values: $T = (40)^2 \times (3 \times 10^{-4}) = 1600 \times 3 \times 10^{-4} = 4800 \times 10^{-4} = 0.48 \ N$.
Comparing this with the standard wave equation $y = A \sin (kx + \omega t)$,we get:
Wave number $k = 1.5 \ m^{-1}$ and angular frequency $\omega = 60 \ rad \ s^{-1}$.
The velocity of the wave is given by $v = \frac{\omega}{k} = \frac{60}{1.5} = 40 \ m \ s^{-1}$.
The velocity of a transverse wave in a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Given $\mu = 3 \times 10^{-4} \ kg \ m^{-1}$.
Rearranging for tension $T$: $T = v^2 \mu$.
Substituting the values: $T = (40)^2 \times (3 \times 10^{-4}) = 1600 \times 3 \times 10^{-4} = 4800 \times 10^{-4} = 0.48 \ N$.
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View Solution98
EasyMCQ
$A$ wire of length $50 \ cm$ and weighing $10 \ g$ is attached to a spring at one end and to a fixed wall at the other end. The spring has a spring constant of $50 \ N/m$ and is stretched by $1 \ cm$. If a wave pulse is produced on the string near the wall,then how much time will it take to reach the spring (in $s$)?
A
$0.1$
B
$0.2$
C
$0.3$
D
$0.4$
Solution
(A) The tension $T$ in the wire is provided by the spring: $T = kx = 50 \ N/m \times 0.01 \ m = 0.5 \ N$.
The mass per unit length $\mu$ of the wire is calculated as: $\mu = \frac{m}{l} = \frac{10 \times 10^{-3} \ kg}{0.5 \ m} = 0.02 \ kg/m$.
The speed $v$ of the wave pulse on the string is given by: $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{0.5}{0.02}} = \sqrt{25} = 5 \ m/s$.
The time $t$ taken by the pulse to travel the length of the wire is: $t = \frac{l}{v} = \frac{0.5 \ m}{5 \ m/s} = 0.1 \ s$.
The mass per unit length $\mu$ of the wire is calculated as: $\mu = \frac{m}{l} = \frac{10 \times 10^{-3} \ kg}{0.5 \ m} = 0.02 \ kg/m$.
The speed $v$ of the wave pulse on the string is given by: $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{0.5}{0.02}} = \sqrt{25} = 5 \ m/s$.
The time $t$ taken by the pulse to travel the length of the wire is: $t = \frac{l}{v} = \frac{0.5 \ m}{5 \ m/s} = 0.1 \ s$.
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View SolutionWaves and Sound — Speed of Mechanical Wave on String (Transverse wave) · Frequently Asked Questions
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