Speed of Mechanical Wave on String (Transverse wave) Questions in English

(A) The speed of sound in a stretched string is given by the formula $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension in the string and $\mu$ is the mass per unit length.
According to Hooke's law,the tension $T$ in a string is directly proportional to the extension $x$,so $T \propto x$.
Substituting this into the velocity formula,we get $v \propto \sqrt{x}$.
Let the initial speed be $v_1 = v$ at extension $x_1 = x$,and the new speed be $v_2$ at extension $x_2 = 1.5x$.
Then,$\frac{v_2}{v_1} = \sqrt{\frac{x_2}{x_1}} = \sqrt{\frac{1.5x}{x}} = \sqrt{1.5}$.
Calculating the value,$\sqrt{1.5} \approx 1.22$.
Therefore,$v_2 = 1.22 \,v$.

(C) transverse wave is a wave in which particles of the medium vibrate in a direction perpendicular to the direction of wave propagation.
Sound waves and compressional waves in a spring are examples of longitudinal waves,where particles vibrate parallel to the direction of wave propagation.
The vibration of a string is a classic example of a transverse wave,as the string particles move up and down while the wave travels along the length of the string.
Therefore,the correct option is $C$.

(D) The given equation of the transverse wave is $y = 0.021 \sin(x + 30t)$.
Comparing this with the standard wave equation $y = A \sin(kx + \omega t)$,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 speed of a transverse wave in a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Given $\mu = 1.3 \times 10^{-4} \ kg/m$.
Substituting the values: $30 = \sqrt{\frac{T}{1.3 \times 10^{-4}}}$.
Squaring both sides: $900 = \frac{T}{1.3 \times 10^{-4}}$.
$T = 900 \times 1.3 \times 10^{-4} = 1170 \times 10^{-4} = 0.117 \ N$.
Rounding to two decimal places,$T \approx 0.12 \ N$.

(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 $\mu = \frac{M}{L} = \frac{0.035 \; kg}{7 \; m} = 0.005 \; kg/m$.
Given tension $T = 60.5 \; N$.
Substituting the values into the formula:
$v = \sqrt{\frac{60.5}{0.005}}$
$v = \sqrt{12100}$
$v = 110 \; m/s$.

(B) The velocity 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.
Since $\mu = \pi r^2 \rho$ (where $r$ is the radius and $\rho$ is the density of the material),the velocity is $v = \sqrt{\frac{T}{\pi r^2 \rho}}$.
Since both wires are made of steel,$\rho$ is constant. Thus,$v \propto \frac{\sqrt{T}}{r}$.
Given: $d_A = 2d_B \implies r_A = 2r_B$ and $T_A = \frac{1}{2}T_B$.
The ratio of velocities is $\frac{v_A}{v_B} = \sqrt{\frac{T_A}{T_B}} \times \frac{r_B}{r_A}$.
Substituting the values: $\frac{v_A}{v_B} = \sqrt{\frac{1}{2}} \times \frac{1}{2} = \frac{1}{\sqrt{2}} \times \frac{1}{2} = \frac{1}{2\sqrt{2}}$.
Therefore,the ratio is $1 : 2\sqrt{2}$.

(B) The linear mass density of the string is $\mu = \frac{M}{L} = \frac{10^{-2}}{0.4} = 2.5 \times 10^{-2}\, kg/m$.
The velocity of the wave pulse in the string is $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{1.6}{2.5 \times 10^{-2}}} = \sqrt{\frac{160}{2.5}} = \sqrt{64} = 8\, m/s$.
For constructive interference,the pulse must return to the starting point in the same phase. When a pulse reflects from a fixed end,it undergoes a phase change of $\pi$. After two reflections (traveling a total distance of $2L$),the pulse undergoes a total phase change of $2\pi$,returning to its original phase.
Therefore,the time taken for the pulse to travel $2L$ is $\Delta t_{\min} = \frac{2L}{v} = \frac{2 \times 0.4}{8} = \frac{0.8}{8} = 0.1\, s$.

(B) The velocity of a transverse wave on a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension in the string and $\mu$ is the linear mass density.
The linear mass density of the string is $\mu = \frac{M}{L}$.
At a distance $x$ from the free end,the tension $T$ in the string is equal to the weight of the portion of the string hanging below that point. The mass of this portion is $m' = \mu x = \left(\frac{M}{L}\right)x$.
Therefore,the tension at distance $x$ is $T = m'g = \left(\frac{M}{L}\right)xg$.
Substituting these values into the velocity formula:
$v = \sqrt{\frac{\left(\frac{M}{L}\right)xg}{\left(\frac{M}{L}\right)}} = \sqrt{gx}$.

(A) 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.
Since the frequency $f$ of the pulse remains constant,the wavelength $\lambda = \frac{v}{f}$ is directly proportional to the velocity $v$. Therefore,$\lambda \propto \sqrt{T}$.
At the lower end of the rope (where the block $m_2$ is attached),the tension $T_1$ is due to the weight of the block: $T_1 = m_2 g$.
At the top of the rope,the tension $T_2$ is due to the weight of both the rope and the block: $T_2 = (m_1 + m_2) g$.
Thus,the ratio of the wavelengths is:
$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{(m_1 + m_2) g}{m_2 g}} = \sqrt{\frac{m_1 + m_2}{m_2}}$.

(D) The speed of a transverse wave on a string is given by the formula:
$v = \sqrt{\frac{T}{\mu}}$ ..........$(1)$
where,$T$ is the tension in the string and $\mu$ is the mass per unit length (linear density).
In this problem:
The tension in the string $(T)$ is due to the weight of the block hanging at the end:
$T = m \times g = 1\, kg \times 10\, m/s^2 = 10\, N$.
The mass per unit length is given as $\mu = 0.001\, kg/m$.
Substituting these values into formula $(1)$:
$v = \sqrt{\frac{10}{0.001}} = \sqrt{10000} = 100\, m/s$.
The time taken $(t)$ by the pulse to travel the length $(L = 1\, m)$ of the string is:
$t = \frac{L}{v} = \frac{1\, m}{100\, m/s} = 0.01\, s$.

(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 rope.
At a height $h$ from the lower end,the tension $T$ in the rope is due to the weight of the portion of the rope below that point.
If the total length of the rope is $L$ and total mass is $M$,then $\mu = \frac{M}{L}$.
The mass of the portion of length $h$ is $m = \mu h$.
Therefore,the tension at height $h$ is $T = mg = \mu h g$.
Substituting this into the speed formula:
$v = \sqrt{\frac{\mu h g}{\mu}} = \sqrt{gh}$.
This shows that $v \propto \sqrt{h}$,which represents a parabolic relationship where $v$ increases with $h$. This corresponds to the curve shown in option $C$.

(C) 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.
Given $v = 100 \, m \, s^{-1}$ and $\mu = 10^{-2} \, kg \, m^{-1}$.
$100 = \sqrt{\frac{T}{10^{-2}}} \implies 100^2 = \frac{T}{10^{-2}} \implies T = 10000 \times 10^{-2} = 100 \, N$.
Since the system is in equilibrium,the tension $T$ in the string is equal to the weight of the hanging mass $M$,so $T = Mg = 100 \, N$. Taking $g = 10 \, m \, s^{-2}$,we get $M = \frac{100}{10} = 10 \, kg$.
For the mass $m$ on the inclined plane,the component of gravity along the plane must balance the tension $T$,so $T = mg \sin(30^o)$.
$100 = m \times 10 \times \frac{1}{2} \implies 100 = 5m \implies m = 20 \, kg$.
Thus,$\frac{m}{M} = \frac{20}{10} = 2$. Looking at the options,$m = 20 \, kg$ is correct.

(C) The speed of the wave in the first string is $V_1 = \sqrt{T/\mu}$.
The speed of the wave in the second string is $V_2 = \sqrt{T/(4\mu)} = V_1/2$.
Since $V_2 < V_1$,the second medium is denser.
When a wave reflects from a denser medium,it undergoes a phase change of $\pi$,which introduces a negative sign.
The reflection coefficient is given by $R = (V_2 - V_1) / (V_2 + V_1)$.
$R = (V_1/2 - V_1) / (V_1/2 + V_1) = (-V_1/2) / (3V_1/2) = -1/3$.
The amplitude of the reflected wave is $A_r = R \times A_i = (-1/3) \times 6 \text{ mm} = -2 \text{ mm}$.
The incident wave is $Y_i = (6 \text{ mm}) \sin(5t + 40x)$. The reflected wave travels in the opposite direction ($-x$ direction),so its phase is $(5t - 40x)$.
Thus,the reflected wave equation is $Y_r = -2 \sin(5t - 40x) \text{ mm}$.

(D) 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.
At a distance $x$ from the lower end,the tension $T(x) = w + \mu x g$,where $w$ is the weight attached at the lower end.
Thus,$v = \sqrt{\frac{w + \mu x g}{\mu}} = \sqrt{\frac{w}{\mu} + xg}$.
As the pulse travels upwards,$x$ increases,so the speed $v$ increases.
Squaring both sides,$v^2 = \frac{w}{\mu} + xg$.
Differentiating with respect to time $t$,$2v \frac{dv}{dt} = g \frac{dx}{dt}$.
Since $\frac{dx}{dt} = v$,we have $2v \frac{dv}{dt} = gv$,which simplifies to $\frac{dv}{dt} = \frac{g}{2}$.
Thus,the pulse travels upwards with a constant acceleration of $\frac{g}{2}$.

(C) Let $x$ be the distance from the bottom end of the string. The linear mass density is $\mu(x) = (\lambda/L)x$.
The tension $T(x)$ at a distance $x$ from the bottom is due to the weight of the segment of length $x$ in the accelerating frame. The effective acceleration is $g_{eff} = g + 2g = 3g$.
The mass of the segment of length $x$ is $m(x) = \int_0^x \mu(x') dx' = \int_0^x (\lambda/L)x' dx' = \frac{\lambda x^2}{2L}$.
The tension $T(x) = m(x) \cdot g_{eff} = \frac{3g\lambda x^2}{2L}$.
The wave velocity $v(x) = \sqrt{T(x)/\mu(x)} = \sqrt{\frac{3g\lambda x^2 / 2L}{\lambda x / L}} = \sqrt{\frac{3gx}{2}}$.
Since $v = dx/dt$,we have $dt = dx / \sqrt{3gx/2} = \sqrt{2/3g} \cdot x^{-1/2} dx$.
Integrating from $x=0$ to $x=L$,the time $t = \int_0^L \sqrt{2/3g} \cdot x^{-1/2} dx = \sqrt{2/3g} \cdot [2x^{1/2}]_0^L = \sqrt{2/3g} \cdot 2\sqrt{L} = \sqrt{8L/3g}$.

(A) The given wave equation is $y = 0.02 \sin \left[ 2\pi \left( \frac{t}{0.04} - \frac{x}{0.50} \right) \right]$.
Comparing this with the standard wave equation $y = A \sin (\omega t - kx)$,we get:
Angular frequency $\omega = \frac{2\pi}{0.04} \ rad/s$ and wave number $k = \frac{2\pi}{0.50} \ rad/m$.
The wave velocity $v$ is given by $v = \frac{\omega}{k} = \frac{2\pi / 0.04}{2\pi / 0.50} = \frac{0.50}{0.04} = 12.5 \ m/s$.
The velocity of a wave on a stretched string is $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Given $\mu = 0.04 \ kg/m$,we have $T = v^2 \mu$.
Substituting the values,$T = (12.5)^2 \times 0.04 = 156.25 \times 0.04 = 6.25 \ N$.

(A) The velocity of a 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.
For a string of length $l$ and mass $m$,$\mu = \frac{m}{l}$.
At a distance $x$ from the lower end,the tension $T$ is due to the weight of the string below that point: $T = \mu x g$.
Substituting this into the velocity formula: $v = \sqrt{\frac{\mu x g}{\mu}} = \sqrt{gx}$.
Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = \sqrt{gx}$.
Separating variables: $x^{-1/2} dx = \sqrt{g} dt$.
Integrating from $x=0$ to $x=l$ and $t=0$ to $t=T_{total}$:
$\int_{0}^{l} x^{-1/2} dx = \int_{0}^{T_{total}} \sqrt{g} dt$.
$[2x^{1/2}]_{0}^{l} = \sqrt{g} T_{total}$.
$2\sqrt{l} = \sqrt{g} T_{total} \implies T_{total} = 2\sqrt{\frac{l}{g}}$.
Given $l = 20 \ m$ and $g = 10 \ ms^{-2}$,we get $T_{total} = 2\sqrt{\frac{20}{10}} = 2\sqrt{2} \ s$.

(C) The velocity of sound in air is given by $V = \sqrt{\frac{\gamma RT}{M}}$.
Since $V \propto \sqrt{T}$,where $T$ is the absolute temperature in Kelvin,we have $V^2 \propto T$.
Option $(A)$ shows velocity vs pressure,which is incorrect as velocity of sound in an ideal gas is independent of pressure.
Option $(B)$ shows $V^2$ vs temperature in $^\circ C$. Since $V^2 \propto (t + 273.15)$,the graph of $V^2$ versus $t$ $(^\circ C)$ should be a straight line with a positive intercept on the $V^2$ axis,not passing through the origin.
Option $(C)$ shows the velocity of a transverse wave in a string,$V = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension. This implies $V \propto \sqrt{T}$,which represents a parabolic curve $V^2 \propto T$. Thus,the graph of $V$ versus $T$ is a parabola opening towards the $T$-axis.
Option $(D)$ shows an inverse relationship,which is incorrect.
Therefore,the correct graph is $(C)$.

(D) In a transverse wave,the elements of the string perform simple harmonic motion $(SHM)$.
For an element in $SHM$,the displacement $y$ is given by $y = A \sin(\omega t + \phi)$.
The velocity is $v = \frac{dy}{dt} = A\omega \cos(\omega t + \phi)$,which is maximum at the mean position $(y = 0)$.
The acceleration is $a = \frac{d^2y}{dt^2} = -\omega^2 y$,which is maximum at the extreme positions $(y = \pm A)$.
Points $P$ and $S$ are at the crests (extreme positions),so the acceleration of the elements at $P$ and $S$ is maximum.
Point $Q$ is at the mean position,so its velocity is maximum,but its displacement is zero only at this instant,not always.
Point $R$ is at the trough (extreme position),where its velocity is zero and its energy is entirely potential.
Therefore,the correct statement is that the acceleration of the element at $S$ is maximum.

(C) The wave equation is given by $y = 2 \sin(20t - 10x)$. Comparing this with the standard wave equation $y = A \sin(\omega t - kx)$,we get angular frequency $\omega = 20 \ rad/s$ and wave number $k = 10 \ m^{-1}$.
The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{20}{10} = 2 \ m/s$.
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.
Linear mass density $\mu = \text{density} \times \text{area} = 1 \ kg/m^3 \times 0.01 \ m^2 = 0.01 \ kg/m$.
The tension $T$ in the string is provided by the weight of the hanging mass $M$,so $T = Mg$.
Substituting these into the speed formula: $2 = \sqrt{\frac{Mg}{0.01}}$.
Squaring both sides: $4 = \frac{Mg}{0.01} \implies 4 = \frac{M \times 10}{0.01} \implies 4 = 1000M$.
Therefore,$M = \frac{4}{1000} = 0.004 \ kg$.

(B) The tension $T$ in the string is due to the mass $M = 1 \ kg$ hanging from it. Since the system is accelerating upwards with $a = 2 \ m/s^2$,the effective acceleration is $g_{eff} = g + a = 10 + 2 = 12 \ m/s^2$.
The tension $T$ at any point in the string (neglecting the mass of the string itself as it is very small compared to $1 \ kg$) is $T = M(g + a) = 1 \times 12 = 12 \ N$.
The speed of the transverse pulse $v$ is given by $v = \sqrt{\frac{T}{\mu}}$,where $\mu = 1.2 \ g/m = 1.2 \times 10^{-3} \ kg/m$.
$v = \sqrt{\frac{12}{1.2 \times 10^{-3}}} = \sqrt{10^4} = 100 \ m/s$.
Since the speed is constant,the time taken $t$ to travel the length $L = 1 \ m$ is $t = \frac{L}{v} = \frac{1}{100} = 0.01 \ s$.

(C) Let $T_1$ and $T_2$ be the tensions in string $1$ and string $2$,respectively. For the mass to be in equilibrium:
$T_1 \sin(37^\circ) + T_2 \sin(53^\circ) = mg = 20 \times 10 = 200\ N$
$T_1 \cos(37^\circ) = T_2 \cos(53^\circ) \Rightarrow T_1 \times (4/5) = T_2 \times (3/5) \Rightarrow T_1 = (3/4) T_2$
Substituting $T_1$ in the vertical equilibrium equation:
$(3/4) T_2 \times (3/5) + T_2 \times (4/5) = 200 \Rightarrow (9/20 + 16/20) T_2 = 200 \Rightarrow (25/20) T_2 = 200 \Rightarrow T_2 = 160\ N$
$T_1 = (3/4) \times 160 = 120\ N$
The lengths of the strings are $L_1 = 10 \sin(53^\circ) = 10 \times (4/5) = 8\ m$ and $L_2 = 10 \sin(37^\circ) = 10 \times (3/5) = 6\ m$.
The wave speed is $v = \sqrt{T/\mu}$. Thus,$v_1 = \sqrt{T_1/\mu}$ and $v_2 = \sqrt{T_2/\mu}$.
The time taken is $t = L/v$. Therefore,$t_1/t_2 = (L_1/v_1) / (L_2/v_2) = (L_1/L_2) \times \sqrt{T_2/T_1}$.
$t_1/t_2 = (8/6) \times \sqrt{160/120} = (4/3) \times \sqrt{4/3} = \frac{4\sqrt{4}}{3\sqrt{3}}$.

(A) The wave equation is given by $Y = A \sin(kx + \omega t)$.
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 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 $\mu = 1.3 \times 10^{-4} \, kg/m$.
Substituting the values: $30 = \sqrt{\frac{T}{1.3 \times 10^{-4}}}$.
Squaring both sides: $900 = \frac{T}{1.3 \times 10^{-4}}$.
$T = 900 \times 1.3 \times 10^{-4} = 1170 \times 10^{-4} = 0.117 \, N$.

(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 $\mu = \frac{m}{l} = \frac{2.5\ kg}{20.0\ m} = 0.125\ kg/m$.
Now,calculate the wave speed $v = \sqrt{\frac{200\ N}{0.125\ kg/m}} = \sqrt{1600} = 40\ m/s$.
The time taken for the disturbance to reach the other end is $t = \frac{l}{v} = \frac{20.0\ m}{40\ m/s} = 0.5\ s$.

(B) Let $L$ be the length of the string and $m$ be its total mass. The linear mass density is $\mu = m/L$.
Consider a point $P$ at a distance $x$ from the bottom end of the string.
The tension $T$ at point $P$ is equal to the weight of the string below it,so $T = (\mu x)g = \frac{m}{L}gx$.
The speed of a transverse pulse in a string is given by $v = \sqrt{T/\mu}$.
Substituting $T$,we get $v = \sqrt{\frac{(\mu x)g}{\mu}} = \sqrt{gx}$.
As the pulse moves downward,$x$ decreases,so the speed $v$ decreases.
To find the rate of change of speed with respect to time,we use $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}$.
Since the pulse moves downward,$\frac{dx}{dt} = -v = -\sqrt{gx}$.
Calculating $\frac{dv}{dx} = \frac{d}{dx}(\sqrt{gx}) = \frac{g}{2\sqrt{gx}}$.
Thus,$a = \left(\frac{g}{2\sqrt{gx}}\right) \cdot (-\sqrt{gx}) = -g/2$.
The negative sign indicates that the speed decreases at a constant rate of $g/2$ as it moves downward.

(A) Let $x$ be the distance from the lower end of the wire. The tension $T$ at a distance $x$ from the lower end is equal to the weight of the wire below that point.
The mass of the wire of length $x$ is $m = \int_0^x \mu(x') dx' = \int_0^x \mu_0 x' dx' = \frac{\mu_0 x^2}{2}$.
Thus,the tension $T$ at distance $x$ is $T = mg = \frac{\mu_0 x^2 g}{2}$.
The speed of the transverse wave at distance $x$ is $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{\mu_0 x^2 g / 2}{\mu_0 x}} = \sqrt{\frac{gx}{2}}$.
Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = \sqrt{\frac{g}{2}} \sqrt{x}$.
Separating variables,$\int_0^{l_0} x^{-1/2} dx = \int_0^t \sqrt{\frac{g}{2}} dt$.
$[2x^{1/2}]_0^{l_0} = \sqrt{\frac{g}{2}} t$.
$2\sqrt{l_0} = \sqrt{\frac{g}{2}} t$.
$t = 2\sqrt{l_0} \sqrt{\frac{2}{g}} = 2\sqrt{\frac{2l_0}{g}} = \sqrt{\frac{8l_0}{g}}$.

(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 mass per unit length.
$\mu = \frac{M}{L}$,where $M$ is the total mass of the wire.
Since density $\rho = \frac{M}{V} = \frac{M}{AL}$,we have $M = \rho AL$.
Substituting $M$ into the expression for $\mu$:
$\mu = \frac{\rho AL}{L} = \rho A$.
Therefore,the speed of the wave is $v = \sqrt{\frac{T}{\rho A}}$.

(C) Given:
Frequency of fundamental mode,$n = 50 \, Hz$
Mass of the wire,$M = 30 \, g = 30 \times 10^{-3} \, kg$
Linear mass density,$\mu = 4 \times 10^{-2} \, kg/m$
First,we find the length of the wire $(L)$:
Since $\mu = M/L$,we have $L = M/\mu$
$L = (30 \times 10^{-3} \, kg) / (4 \times 10^{-2} \, kg/m) = 0.75 \, m$
For the fundamental mode of vibration in a string fixed at both ends:
$L = \lambda / 2$
$\lambda = 2L = 2 \times 0.75 \, m = 1.5 \, m$
The speed of the transverse wave $(v)$ is given by:
$v = n \lambda$
$v = 50 \, Hz \times 1.5 \, m = 75 \, m/s$

(A) The speed of the wave in the rope is given by $v = \frac{L}{t} = \frac{43\, m}{1.4\, s} \approx 30.71\, m/s$.
The linear mass density $\mu$ of the rope is $\mu = \frac{m}{L} = \frac{5.0\, kg}{43\, m} \approx 0.116\, kg/m$.
The wave speed in a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension.
Rearranging for tension: $T = v^2 \mu = \left(\frac{43}{1.4}\right)^2 \times \frac{5.0}{43}$.
$T = \frac{43^2}{1.96} \times \frac{5}{43} = \frac{43 \times 5}{1.96} = \frac{215}{1.96} \approx 109.69\, N$.
Rounding to the nearest whole number,the tension is approximately $110\, N$.

(B) The velocity of the transverse wave in the string is given by $v = \frac{L}{t}$,where $L = 20 \, m$ and $t = 0.5 \, s$.
$v = \frac{20}{0.5} = 40 \, m/s$.
The velocity of a transverse wave in a string is also given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Linear mass density $\mu = \frac{m}{L} = \frac{2.5 \, kg}{20 \, m} = 0.125 \, kg/m$.
Squaring the velocity formula,we get $v^2 = \frac{T}{\mu}$,which implies $T = v^2 \times \mu$.
Substituting the values: $T = (40)^2 \times 0.125 = 1600 \times 0.125 = 200 \, N$.

(C) The given wave equation is $y = 0.03 \sin(450t - 9x)$.
Comparing this with the standard wave equation $y = A \sin(\omega t - kx)$,we get angular frequency $\omega = 450 \, rad/s$ and wave number $k = 9 \, m^{-1}$.
The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{450}{9} = 50 \, m/s$.
The linear mass density $\mu = 5 \, g/m = 5 \times 10^{-3} \, kg/m$.
The speed of a wave on a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension.
Squaring both sides,we get $v^2 = \frac{T}{\mu}$,so $T = \mu v^2$.
Substituting the values,$T = (5 \times 10^{-3} \, kg/m) \times (50 \, m/s)^2$.
$T = 5 \times 10^{-3} \times 2500 = 5 \times 2.5 = 12.5 \, N$.

(A) Let the mass of the string be $m$ and its length be $l$. The linear mass density is $\mu = \frac{m}{l}$.
The tension at any point in the string is equal to the weight of the portion of the string below that point.
For point $A$,the length of the string below it is $\frac{l}{4}$. Thus,the mass below $A$ is $m_A = \mu \times \frac{l}{4} = \frac{m}{4}$.
The tension at $A$ is $T_A = m_A g = \frac{mg}{4}$.
The speed of the wave at $A$ is $v_A = \sqrt{\frac{T_A}{\mu}} = \sqrt{\frac{mg/4}{m/l}} = \sqrt{\frac{gl}{4}}$.
For point $B$,the length of the string below it is $\frac{3l}{4}$. Thus,the mass below $B$ is $m_B = \mu \times \frac{3l}{4} = \frac{3m}{4}$.
The tension at $B$ is $T_B = m_B g = \frac{3mg}{4}$.
The speed of the wave at $B$ is $v_B = \sqrt{\frac{T_B}{\mu}} = \sqrt{\frac{3mg/4}{m/l}} = \sqrt{\frac{3gl}{4}}$.
Comparing $v_B$ and $v_A$:
$v_B = \sqrt{3} \times \sqrt{\frac{gl}{4}} = \sqrt{3} v_A$.

(B) Given: Length of the string $L = 1\,m$,Frequency $f = 300\,Hz$,Number of segments $n = 3$.
For a string vibrating in $n$ segments,the length of each segment is $\lambda/2 = L/n$.
Therefore,the wavelength $\lambda = 2L/n$.
Substituting the values: $\lambda = 2 \times 1 / 3 = 2/3\,m$.
The speed of the transverse wave is given by $v = f \lambda$.
$v = 300 \times (2/3) = 200\,m/s$.

(C) $1$. For the velocity of sound in air, $v = \sqrt{\frac{\gamma P}{\rho}}$. Since $\rho = \frac{PM}{RT}$, we have $v = \sqrt{\frac{\gamma RT}{M}}$. At constant temperature $(T)$, $v$ is independent of pressure $(P)$. Thus, the graph in option $A$ is incorrect as it shows $v$ increasing with $P$.
$2$. The velocity of sound in air is $v = \sqrt{\frac{\gamma R(T_c + 273)}{M}}$. Thus, $v^2 = \frac{\gamma R}{M}(T_c + 273)$. This is a linear equation of the form $y = mx + c$ where $y = v^2$ and $x = T_c$. The graph in option $B$ shows a line passing through the origin, which is incorrect because at $T_c = 0^\circ C$, $v^2 \neq 0$.
$3$. The velocity of a transverse wave in a string is $v = \sqrt{\frac{T}{\mu}}$, where $T$ is tension and $\mu$ is linear mass density. This implies $v^2 = \frac{1}{\mu}T$, or $v = \frac{1}{\sqrt{\mu}}\sqrt{T}$. This represents a parabolic relationship between $v$ and $T$. The graph in option $C$ correctly shows this parabolic dependence.
Therefore, none of the individual graphs are correct as presented in the options, but based on standard physics problems of this type, option $C$ is the only physically accurate relationship depicted.

(B) The speed of a transverse wave pulse in a stretched string is given by the formula $v = \sqrt{T / \mu}$,where $T$ is the tension in the rope and $\mu$ is the mass per unit length of the rope.
For a heavy rope suspended from a rigid support,the tension $T$ at any point at a distance $x$ from the lower end is due to the weight of the rope below that point. As the pulse travels upwards from the lower end to the upper end,the length of the rope below the pulse increases,which means the weight of the rope below the pulse increases.
Consequently,the tension $T$ at the position of the pulse increases as it moves upwards. Since $v = \sqrt{T / \mu}$ and $\mu$ is constant,the speed $v$ of the pulse increases as it travels towards the support.

(C) Given: Length of the wire,$L = 10 \, m$.
Mass of the wire,$M = 5 \, g = 5 \times 10^{-3} \, kg$.
Tension in the wire,$T = 80 \, N$.
First,calculate the linear mass density (mass per unit length) of the wire,denoted by $\mu$:
$\mu = \frac{M}{L} = \frac{5 \times 10^{-3} \, kg}{10 \, m} = 5 \times 10^{-4} \, kg \cdot m^{-1}$.
The speed $v$ of a transverse wave on a stretched string is given by the formula:
$v = \sqrt{\frac{T}{\mu}}$.
Substituting the values:
$v = \sqrt{\frac{80 \, N}{5 \times 10^{-4} \, kg \cdot m^{-1}}} = \sqrt{16 \times 10^4} \, m \cdot s^{-1}$.
$v = 400 \, m \cdot s^{-1}$.

(B) The speed of a transverse wave in a string is given by $v = \sqrt{T/\mu} = f\lambda$,where $T$ is the tension,$\mu$ is the linear mass density,and $f$ is the frequency.
Since the frequency $f$ remains constant as the pulse travels,we have $\lambda \propto \sqrt{T}$.
At the bottom of the rope,the tension $T_1$ is due to the block of mass $2\,kg$,so $T_1 = 2g$.
At the top of the rope,the tension $T_2$ is due to the combined mass of the rope $(6\,kg)$ and the block $(2\,kg)$,so $T_2 = (6 + 2)g = 8g$.
Using the proportionality $\lambda \propto \sqrt{T}$,we get $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values,$\frac{\lambda_2}{0.06} = \sqrt{\frac{8g}{2g}} = \sqrt{4} = 2$.
Therefore,$\lambda_2 = 0.06 \times 2 = 0.12\,m$.

(A) Let $T$ be the tension at a point at a distance $x$ from the free end.
The mass of the string segment of length $x$ is $m_x = \frac{M}{L} x$.
The tension $T$ at this point is equal to the weight of the string segment below it:
$T = m_x \cdot g = \frac{M}{L} x g$.
The linear mass density of the string is $\mu = \frac{M}{L}$.
The velocity of a transverse wave on a string is given by $v = \sqrt{\frac{T}{\mu}}$.
Substituting the values of $T$ and $\mu$:
$v = \sqrt{\frac{(M/L) x g}{M/L}} = \sqrt{gx}$.

(D) The speed of a wave in a string is given by $v = \sqrt{T/\mu}$,where $T$ is the tension and $\mu$ is the linear mass density. Since both $T$ and $\mu$ are uniform for the string,the speed $v$ is constant for all waves traveling in it.
However,velocity is a vector quantity that includes both speed and direction. Two waves can travel in the same string with the same speed $v$ but in opposite directions (e.g.,one moving in the $+x$ direction and the other in the $-x$ direction).
Thus,the velocities of these two waves would be $+v$ and $-v$,which are different. Therefore,the Assertion that they cannot have different velocities is incorrect.
The Reason correctly states that the speed depends only on the elastic and inertial properties of the string,which are uniform,but the Assertion itself is false.

(B) The velocity 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.
From this relation,we see that $v \propto \sqrt{T}$.
Let the initial tension be $T_1 = 2.06 \times 10^{4} \; N$ and the initial velocity be $v_1 = v$.
Let the final tension be $T_2 = T$ and the final velocity be $v_2 = v/2$.
Using the proportionality $v \propto \sqrt{T}$,we can write: $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the given values: $\frac{v/2}{v} = \sqrt{\frac{T}{2.06 \times 10^{4}}}$.
$\frac{1}{2} = \sqrt{\frac{T}{2.06 \times 10^{4}}}$.
Squaring both sides: $\frac{1}{4} = \frac{T}{2.06 \times 10^{4}}$.
$T = \frac{2.06 \times 10^{4}}{4} = 0.515 \times 10^{4} = 5.15 \times 10^{3} \; N$.

(C) The mass per unit length (linear mass density) of the wire is given by $\mu = \frac{m}{L}$.
Given,mass $m = 5.0 \times 10^{-3} \; kg$ and length $L = 0.72 \; m$.
$\mu = \frac{5.0 \times 10^{-3} \; kg}{0.72 \; m} \approx 6.944 \times 10^{-3} \; kg/m$.
The tension in the wire is $T = 60 \; N$.
The speed of a transverse wave on a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$.
Substituting the values: $v = \sqrt{\frac{60}{6.944 \times 10^{-3}}} = \sqrt{8640.55} \approx 92.95 \; m/s$.
Rounding to the nearest whole number,the speed is $93 \; m/s$.

(D) Mass of the string,$M = 2.50 \; kg$.
Tension in the string,$T = 200 \; N$.
Length of the string,$l = 20.0 \; m$.
Mass per unit length,$\mu = \frac{M}{l} = \frac{2.50}{20} = 0.125 \; kg \; m^{-1}$.
The velocity $(v)$ of the transverse wave in the string is given by the relation:
$v = \sqrt{\frac{T}{\mu}}$
$v = \sqrt{\frac{200}{0.125}} = \sqrt{1600} = 40 \; m/s$.
Therefore,the time taken by the disturbance to reach the other end is:
$t = \frac{l}{v} = \frac{20}{40} = 0.50 \; s$.

(B) Length of the steel wire,$l = 12.0 \;m$.
Mass of the steel wire,$m = 2.10 \;kg$.
Velocity of the transverse wave,$v = 343 \;m s^{-1}$.
Mass per unit length,$\mu = \frac{m}{l} = \frac{2.10}{12.0} = 0.175 \;kg \;m^{-1}$.
For tension $T$,the velocity of the transverse wave is given by the relation: $v = \sqrt{\frac{T}{\mu}}$.
Rearranging for $T$,we get $T = v^2 \mu$.
Substituting the values: $T = (343)^2 \times 0.175 = 117649 \times 0.175 = 20588.575 \;N$.
Rounding to three significant figures,$T \approx 2.06 \times 10^{4} \;N$.

(N/A) The speed of transverse waves on a string is determined by two factors: $(i)$ the linear mass density or mass per unit length $\mu$ and $(ii)$ the tension $T$.
The linear mass density $\mu$ of a string is the mass $m$ of the string divided by its length $l$. Therefore,its dimension is $[M^1 L^{-1}]$. The tension $T$ has the dimension of force,namely $[M^1 L^1 T^{-2}]$.
We have to combine $\mu$ and $T$ in such a way as to generate $v$ [dimension $(L T^{-1})$].
It can be seen that the ratio $\frac{T}{\mu}$ has the dimension $[L^2 T^{-2}]$.
$\left[\frac{T}{\mu}\right] = \frac{[M^1 L^1 T^{-2}]}{[M^1 L^{-1}]} = [L^2 T^{-2}]$
Therefore,if $v$ depends only on $T$ and $\mu$,the relation between them must be:
$v = C \sqrt{\frac{T}{\mu}}$
Here $C$ is a dimensionless constant,and it is found that $C = 1$.
The speed of transverse waves on a stretched string is therefore given by:
$v = \sqrt{\frac{T}{\mu}}$

(N/A) The speed $v$ of a transverse wave on a stretched string is given by the formula:
$v = \sqrt{\frac{T}{\mu}}$
Where:
$T$ is the tension in the string (measured in Newtons,$N$),
$\mu$ is the linear mass density of the string (mass per unit length,measured in $kg/m$).

(A) The speed of a wave in a stretched wire is given by the formula $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
From the formula,we can see that $v \propto \sqrt{T}$.
Let the initial tension be $T_1$ and the initial speed be $v_1$. Let the final tension be $T_2 = 4T_1$ and the final speed be $v_2$.
Using the proportionality,we have $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values,we get $\frac{v_2}{v_1} = \sqrt{\frac{4T_1}{T_1}} = \sqrt{4} = 2$.
Therefore,$v_2 = 2v_1$. The speed of the wave will become double.

(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 $\mu = \frac{m}{L}$,where $m$ is the mass and $L$ is the length of the wire.
If the length $L$ is halved $(L' = L/2)$ and mass $m$ remains constant,the new linear mass density $\mu'$ becomes $\mu' = \frac{m}{L/2} = 2\mu$.
Assuming the tension $T$ remains constant,the new speed $v'$ is $v' = \sqrt{\frac{T}{\mu'}} = \sqrt{\frac{T}{2\mu}} = \frac{1}{\sqrt{2}} v$.
However,if the question implies the tension $T$ changes due to the change in length (as in a stretched string),we must consider the elastic properties. In standard textbook problems of this type,if only length is changed while keeping mass constant,the speed changes by a factor of $1/\sqrt{2}$.
Given the provided solution logic suggests $v_2 = 2v_1$,it assumes the tension $T$ increases by a factor of $4$. If we strictly follow the physics of a wire with constant mass,the speed becomes $1/\sqrt{2}$ times the original speed.

(A) The frequency $f$ of a vibrating string is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length,$T$ is the tension,and $\mu$ is the linear mass density.
Linear mass density $\mu = \text{Area} \times \text{Density} = (\pi r^2) \rho$.
Substituting this into the frequency formula: $f = \frac{1}{2L} \sqrt{\frac{T}{\pi r^2 \rho}}$.
Since $L$,$T$,and $\rho$ are constant,we have $f \propto \frac{1}{r}$.
Let the initial radius be $r_1 = r$ and the new radius be $r_2 = 3r$.
The ratio of the new frequency $f_2$ to the initial frequency $f_1$ is $\frac{f_2}{f_1} = \frac{r_1}{r_2} = \frac{r}{3r} = \frac{1}{3}$.
Thus,the frequency changes by a factor of $1/3$.

(A) 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 linear mass density (mass per unit length).
Given:
Length $L = 12 \ m$
Mass $M = 2.10 \ kg$
Tension $T = 2.06 \times 10^4 \ N$
Linear mass density $\mu = \frac{M}{L} = \frac{2.10}{12} \ kg/m = 0.175 \ kg/m$
Substituting the values into the formula:
$v = \sqrt{\frac{2.06 \times 10^4}{0.175}}$
$v = \sqrt{117714.28} \approx 343.09 \ m/s$
Rounding to the nearest whole number,the speed is $343 \ m/s$.

(B) 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 mass per unit length.
Since the frequency $f$ of the wavetrain remains constant,$v = f \lambda$,which implies $v \propto \lambda$.
At the lower end (bottom),the tension $T_1$ is due to the block of mass $m = 2 \, kg$:
$T_1 = mg = 2g$.
At the top of the rope,the tension $T_2$ is due to the block and the entire mass of the rope $M = 6 \, kg$:
$T_2 = (M + m)g = (6 + 2)g = 8g$.
Using the relation $\frac{\lambda_2}{\lambda_1} = \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$:
$\lambda_2 = \lambda_1 \sqrt{\frac{8g}{2g}} = 6 \times \sqrt{4} = 6 \times 2 = 12 \, cm$.

(C) The given string wave equation is $y=0.002 \sin (300 t-15 x)$.
Comparing this with the standard wave equation $y=A \sin (\omega t - kx)$,we get:
$\omega = 300 \ rad/s$
$k = 15 \ rad/m$
The wave speed $v$ is given by $v = \frac{\omega}{k}$.
$v = \frac{300}{15} = 20 \ m/s$.
The speed of a wave in a 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 $v^2 = \frac{T}{\mu}$.
$T = \mu v^2$.
Substituting the values,$T = 0.1 \times (20)^2$.
$T = 0.1 \times 400 = 40 \ N$.