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Transverse Stationary Waves and Sonometer Questions in English

Class 11 Physics · Waves and Sound · Transverse Stationary Waves and Sonometer

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Showing 50 of 225 questions in English

101
MediumMCQ
The length of the string of a musical instrument is $90 \;cm$ and has a fundamental frequency of $120 \;Hz$. Where (in $cm$) should it be pressed to produce a fundamental frequency of $180 \;Hz$?
A
$80$
B
$75$
C
$60$
D
$45$

Solution

(C) The fundamental frequency of a stretched string is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.
Since the tension $T$ and mass per unit length $\mu$ are constant,the frequency is inversely proportional to the length of the string: $n \propto \frac{1}{L}$.
Therefore,$n_1 L_1 = n_2 L_2$.
Given $n_1 = 120 \;Hz$,$L_1 = 90 \;cm$,and $n_2 = 180 \;Hz$.
Substituting the values: $120 \times 90 = 180 \times L_2$.
$L_2 = \frac{120 \times 90}{180} = \frac{10800}{180} = 60 \;cm$.
Thus,the string should be pressed at $60 \;cm$ from one end to produce the required frequency.
102
DifficultMCQ
$A$ wire having a linear mass density $9.0 \times 10^{-4} \; \text{kg/m}$ is stretched between two rigid supports with a tension of $900 \; \text{N}$. The wire resonates at a frequency of $500 \; \text{Hz}$. The next higher frequency at which the same wire resonates is $550 \; \text{Hz}$. The length of the wire is $...... \; \text{m}$.
A
$50$
B
$100$
C
$10$
D
$2$

Solution

(C) The wave speed $v$ in the wire is given by $v = \sqrt{\frac{T}{\mu}}$.
Given $T = 900 \; \text{N}$ and $\mu = 9.0 \times 10^{-4} \; \text{kg/m}$.
$v = \sqrt{\frac{900}{9.0 \times 10^{-4}}} = \sqrt{10^6} = 1000 \; \text{m/s}$.
The resonant frequencies of a string fixed at both ends are given by $f_n = \frac{nv}{2L}$,where $n = 1, 2, 3, \dots$.
The difference between two consecutive resonant frequencies is $\Delta f = f_{n+1} - f_n = \frac{(n+1)v}{2L} - \frac{nv}{2L} = \frac{v}{2L}$.
Given $\Delta f = 550 \; \text{Hz} - 500 \; \text{Hz} = 50 \; \text{Hz}$.
Therefore,$\frac{v}{2L} = 50 \; \text{Hz}$.
Substituting $v = 1000 \; \text{m/s}$,we get $\frac{1000}{2L} = 50$.
$2L = \frac{1000}{50} = 20$.
$L = 10 \; \text{m}$.
103
MediumMCQ
$A$ wire of length $0.3\,m$,stretched between rigid supports,has its $n^{\text{th}}$ and $(n+1)^{\text{th}}$ harmonics at $400\,Hz$ and $450\,Hz$,respectively. If the tension in the string is $2700\,N$,its linear mass density is ......... $kg/m$.
A
$1.5$
B
$6$
C
$9$
D
$3$

Solution

(D) The frequency of the $n^{\text{th}}$ harmonic is given by $f_n = \frac{nv}{2L} = 400\,Hz$.
The frequency of the $(n+1)^{\text{th}}$ harmonic is $f_{n+1} = \frac{(n+1)v}{2L} = 450\,Hz$.
The difference between consecutive harmonics is $\Delta f = f_{n+1} - f_n = \frac{v}{2L} = 450 - 400 = 50\,Hz$.
Given length $L = 0.3\,m$,we have $\frac{v}{2(0.3)} = 50$,which implies $v = 50 \times 0.6 = 30\,m/s$.
The wave speed $v$ is related to tension $T$ and linear mass density $\mu$ by $v = \sqrt{\frac{T}{\mu}}$.
Substituting the values: $30 = \sqrt{\frac{2700}{\mu}}$.
Squaring both sides: $900 = \frac{2700}{\mu}$.
Therefore,$\mu = \frac{2700}{900} = 3\,kg/m$.
104
EasyMCQ
If the fundamental frequency of a string is $220 \, cps$,the frequency of the fifth harmonic will be ......... $cps$.
A
$44$
B
$55$
C
$1100$
D
$440$

Solution

(C) The fundamental frequency of a string is given as $f_1 = 220 \, cps$.
In a vibrating string,the frequency of the $n^{th}$ harmonic is given by the formula $f_n = n \times f_1$.
For the fifth harmonic,$n = 5$.
Therefore,the frequency of the fifth harmonic is $f_5 = 5 \times 220 \, cps = 1100 \, cps$.
Thus,the correct option is $C$.
105
MediumMCQ
The tension in a wire is decreased by $19 \%$. The percentage decrease in frequency will be ......... $\%$
A
$0.19$
B
$10$
C
$19$
D
$0.9$

Solution

(B) The fundamental frequency $f$ of a stretched wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension,$l$ is the length,and $\mu$ is the linear mass density.
From this relation,$f \propto \sqrt{T}$.
Let the initial tension be $T_1 = T$ and the final tension be $T_2 = T - 0.19T = 0.81T$.
The new frequency $f^{\prime}$ is given by $f^{\prime} = \sqrt{\frac{T_2}{T_1}} f = \sqrt{0.81} f = 0.9f$.
The decrease in frequency is $\Delta f = f - f^{\prime} = f - 0.9f = 0.1f$.
The percentage decrease in frequency is $\frac{\Delta f}{f} \times 100 = 0.1 \times 100 = 10 \%$.
106
MediumMCQ
$A$ $12 \,m$ long vibrating string has a wave speed of $48 \,m/s$. At what frequencies will it resonate in $cps$?
A
$2$
B
$4$
C
$6$
D
All of these

Solution

(D) The frequency of a vibrating string fixed at both ends is given by the formula $f_n = \frac{n v}{2 l}$,where $n = 1, 2, 3, \dots$ is the harmonic number.
Given: length $l = 12 \,m$ and wave speed $v = 48 \,m/s$.
Substituting the values: $f_n = \frac{n \times 48}{2 \times 12} = \frac{48 n}{24} = 2n \,Hz$.
For $n=1$,$f_1 = 2 \,Hz$.
For $n=2$,$f_2 = 4 \,Hz$.
For $n=3$,$f_3 = 6 \,Hz$.
Since the string can resonate at any integer multiple of the fundamental frequency $(2 \,Hz)$,all the given options $(2, 4, 6)$ are possible resonant frequencies.
Therefore,the correct option is $(d)$.
107
EasyMCQ
$A$ certain string will resonate at several frequencies,the lowest of which is $200 \,Hz$. What are the next three higher frequencies at which it resonates?
A
$400, 600, 800 \,Hz$
B
$300, 400, 500 \,Hz$
C
$100, 150, 200 \,Hz$
D
$200, 250, 300 \,Hz$

Solution

(A) The resonant frequencies of a string fixed at both ends are given by the formula $f_n = n \times f_1$,where $n = 1, 2, 3, \dots$ and $f_1$ is the fundamental frequency.
Given that the lowest resonant frequency (fundamental frequency) is $f_1 = 200 \,Hz$.
The next three higher resonant frequencies correspond to $n = 2, 3,$ and $4$.
For $n = 2$,$f_2 = 2 \times 200 = 400 \,Hz$.
For $n = 3$,$f_3 = 3 \times 200 = 600 \,Hz$.
For $n = 4$,$f_4 = 4 \times 200 = 800 \,Hz$.
Therefore,the next three higher frequencies are $400 \,Hz, 600 \,Hz,$ and $800 \,Hz$.
108
MediumMCQ
The string of a violin has a frequency of $440 \,cps$. If the violin string is shortened by one fifth,its frequency will be changed to ........... $cps$.
A
$440$
B
$880$
C
$550$
D
$2200$

Solution

(D) The fundamental frequency of a vibrating string is given by $f = \frac{v}{2l}$,where $v$ is the wave speed and $l$ is the length of the string.
Given initial frequency $f_1 = 440 \,cps$ for length $l_1 = l$.
If the string is shortened by one-fifth,the new length becomes $l_2 = l - \frac{l}{5} = \frac{4l}{5}$.
However,the question states the string is shortened *by* one-fifth,implying the new length is $l_2 = \frac{4l}{5}$. If the interpretation is that the string is shortened *to* one-fifth of its original length,$l_2 = \frac{l}{5}$.
Given the options,we calculate for $l_2 = \frac{l}{5}$:
$f_2 = \frac{v}{2l_2} = \frac{v}{2(l/5)} = 5 \times \frac{v}{2l} = 5 \times f_1$.
$f_2 = 5 \times 440 = 2200 \,cps$.
109
MediumMCQ
$A$ wire of length $1 \,m$ under a certain initial tension emits a sound of fundamental frequency $256 \,Hz$. When the tension is increased by $1 \,kg \,wt$,the frequency of the fundamental note increases to $320 \,Hz$. The initial tension is ........... $kg \,wt$.
A
$3/4$
B
$4/3$
C
$16/9$
D
$20/9$

Solution

(C) The fundamental frequency of a stretched wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Let the initial tension be $T$. Then,$256 = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$.
When the tension is increased by $1 \,kg \,wt$,the new tension is $(T+1) \,kg \,wt$. The new frequency is $320 = \frac{1}{2l} \sqrt{\frac{T+1}{\mu}}$.
Dividing the two equations: $\frac{320}{256} = \sqrt{\frac{T+1}{T}}$.
Simplifying the ratio: $\frac{320}{256} = \frac{5}{4}$.
Squaring both sides: $\frac{25}{16} = \frac{T+1}{T}$.
$25T = 16(T+1) \implies 25T = 16T + 16$.
$9T = 16 \implies T = \frac{16}{9} \,kg \,wt$.
110
MediumMCQ
$A$ tuning fork vibrating with a sonometer having a wire of length $20 \,cm$ produces $5$ beats per second. The beat frequency does not change if the length of the wire is changed to $21 \,cm$. The frequency of the tuning fork must be ............ $Hz$.
A
$200$
B
$210$
C
$205$
D
$215$

Solution

(C) Let the frequency of the tuning fork be $f$ and the frequency of the sonometer wire be $n = \frac{v}{2L}$.
Given that the beat frequency is $5 \,Hz$,we have $|f - n| = 5$.
For $L_1 = 0.20 \,m$,$n_1 = \frac{v}{2 \times 0.20} = \frac{v}{0.4}$.
For $L_2 = 0.21 \,m$,$n_2 = \frac{v}{2 \times 0.21} = \frac{v}{0.42}$.
Since $L_2 > L_1$,$n_2 < n_1$.
If the beat frequency remains $5 \,Hz$,the tuning fork frequency $f$ must lie between $n_1$ and $n_2$ or be such that the difference remains constant.
Case $1$: $n_1 - f = 5 \implies f = n_1 - 5$.
Case $2$: $f - n_2 = 5 \implies f = n_2 + 5$.
Equating the two: $n_1 - 5 = n_2 + 5 \implies n_1 - n_2 = 10$.
Substituting $n_1 = \frac{v}{0.4}$ and $n_2 = \frac{v}{0.42}$:
$\frac{v}{0.4} - \frac{v}{0.42} = 10 \implies v \left( \frac{0.42 - 0.4}{0.168} \right) = 10 \implies v \left( \frac{0.02}{0.168} \right) = 10 \implies v = \frac{1.68}{0.02} = 84 \,m/s$.
Now,$n_1 = \frac{84}{0.4} = 210 \,Hz$.
Since $f = n_1 - 5 = 210 - 5 = 205 \,Hz$.
111
EasyMCQ
$A$ second harmonic has to be generated in a string of length $l$ stretched between two rigid supports. The points where the string has to be plucked and touched are respectively
A
$\frac{l}{4}, \frac{l}{2}$
B
$\frac{l}{4}, \frac{3l}{4}$
C
$\frac{l}{2}, \frac{l}{2}$
D
$\frac{l}{2}, \frac{3l}{4}$

Solution

(A) In the $2^{\text{nd}}$ harmonic,the string vibrates in two loops,meaning there is one node in the middle of the string at $x = \frac{l}{2}$ and two antinodes at $x = \frac{l}{4}$ and $x = \frac{3l}{4}$.
To generate the $2^{\text{nd}}$ harmonic,we must force a node at the center by touching the string at $x = \frac{l}{2}$.
To excite the vibration,we must pluck the string at an antinode position. The available antinode positions are $\frac{l}{4}$ or $\frac{3l}{4}$.
Comparing this with the given options,the correct pair for (plucking point,touching point) is $(\frac{l}{4}, \frac{l}{2})$.
Thus,the correct option is $A$.
Solution diagram
112
MediumMCQ
$A$ uniform string resonates with a tuning fork at a maximum tension of $32 \,N$. If it is divided into two segments by placing a wedge at a distance one-fourth of the length from one end,then to resonate with the same frequency,the maximum value of tension for the string will be ........... $N$.
A
$2$
B
$4$
C
$8$
D
$16$

Solution

(A) The frequency of a vibrating string is given by $f = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$,where $n$ is the number of loops,$L$ is the length,$T$ is the tension,and $\mu$ is the linear mass density.
Initially,the whole string resonates in its fundamental mode $(n=1)$: $f = \frac{1}{2L} \sqrt{\frac{T_0}{\mu}}$,where $T_0 = 32 \,N$.
When a wedge is placed at $L/4$ from one end,the string is divided into two segments of lengths $L_1 = L/4$ and $L_2 = 3L/4$.
For the segment of length $L_1 = L/4$,the fundamental frequency is $f_1 = \frac{1}{2(L/4)} \sqrt{\frac{T_1}{\mu}} = \frac{2}{L} \sqrt{\frac{T_1}{\mu}}$.
Equating $f = f_1$: $\frac{1}{2L} \sqrt{\frac{T_0}{\mu}} = \frac{2}{L} \sqrt{\frac{T_1}{\mu}} \implies \frac{1}{2} \sqrt{T_0} = 2 \sqrt{T_1} \implies \sqrt{T_0} = 4 \sqrt{T_1} \implies T_1 = \frac{T_0}{16} = \frac{32}{16} = 2 \,N$.
For the segment of length $L_2 = 3L/4$,the fundamental frequency is $f_2 = \frac{1}{2(3L/4)} \sqrt{\frac{T_2}{\mu}} = \frac{2}{3L} \sqrt{\frac{T_2}{\mu}}$.
Equating $f = f_2$: $\frac{1}{2L} \sqrt{\frac{T_0}{\mu}} = \frac{2}{3L} \sqrt{\frac{T_2}{\mu}} \implies \frac{1}{2} \sqrt{T_0} = \frac{2}{3} \sqrt{T_2} \implies \sqrt{T_2} = \frac{3}{4} \sqrt{T_0} \implies T_2 = \frac{9}{16} T_0 = \frac{9}{16} \times 32 = 18 \,N$.
The question asks for the maximum value of tension,which is $18 \,N$. However,since $18 \,N$ is not in the options,we check the segment $L_1$,which gives $2 \,N$.
113
DifficultMCQ
The frequency of a sonometer wire is $100 \ Hz$. When the weights producing the tension are completely immersed in water,the frequency becomes $80 \ Hz$,and on immersing the weights in a certain liquid,the frequency becomes $60 \ Hz$. The specific gravity of the liquid is:
A
$1.42$
B
$1.77$
C
$1.82$
D
$1.21$

Solution

(B) The frequency of a sonometer wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension. Since $T = mg$,we have $f \propto \sqrt{T} \propto \sqrt{mg_{eff}}$.
When the weight is immersed in a liquid of density $\rho_l$,the effective gravity is $g' = g(1 - \frac{\rho_l}{\rho_m})$,where $\rho_m$ is the density of the weight.
Thus,$f \propto \sqrt{1 - \frac{\rho_l}{\rho_m}}$.
In air,$f_0 = 100 \ Hz$.
In water $(\rho_w = 1 \ g/cm^3)$,$f_w = 80 \ Hz \implies \frac{f_w}{f_0} = \frac{80}{100} = 0.8$.
So,$0.8 = \sqrt{1 - \frac{\rho_w}{\rho_m}} \implies 0.64 = 1 - \frac{\rho_w}{\rho_m} \implies \frac{\rho_w}{\rho_m} = 0.36$.
In the liquid,$f_l = 60 \ Hz \implies \frac{f_l}{f_0} = \frac{60}{100} = 0.6$.
So,$0.6 = \sqrt{1 - \frac{\rho_l}{\rho_m}} \implies 0.36 = 1 - \frac{\rho_l}{\rho_m} \implies \frac{\rho_l}{\rho_m} = 0.64$.
Dividing the two ratios: $\frac{\rho_l}{\rho_w} = \frac{0.64}{0.36} = \frac{64}{36} = \frac{16}{9} \approx 1.77$.
Since the specific gravity is $\frac{\rho_l}{\rho_w}$,the value is $1.77$.
114
EasyMCQ
$A$ guitar string of length $90\,cm$ vibrates with a fundamental frequency of $120\,Hz.$ The length of the string producing a fundamental frequency of $180\,Hz$ will be $...........cm$.
A
$60$
B
$59$
C
$58$
D
$57$

Solution

(A) For a string fixed at both ends,the fundamental frequency $f$ is given by $f = \frac{v}{2\ell}$,where $v$ is the wave speed and $\ell$ is the length of the string.
Since $v$ depends on the tension and linear mass density of the string,which remain constant,we have $f \propto \frac{1}{\ell}$.
Therefore,$f_1 \ell_1 = f_2 \ell_2$.
Given $f_1 = 120\,Hz$,$\ell_1 = 90\,cm$,and $f_2 = 180\,Hz$.
Substituting the values: $120 \times 90 = 180 \times \ell_2$.
$\ell_2 = \frac{120 \times 90}{180} = \frac{10800}{180} = 60\,cm$.
115
MediumMCQ
$A$ wire of density $8 \times 10^3\,kg/m^3$ is stretched between two clamps $0.5\,m$ apart. The extension developed in the wire is $3.2 \times 10^{-4}\,m$. If Young's modulus $Y = 8 \times 10^{10}\,N/m^2$,the fundamental frequency of vibration in the wire will be $......\,Hz$.
A
$80$
B
$60$
C
$40$
D
$20$

Solution

(A) The fundamental frequency of a stretched wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.
Here,$T$ is the tension and $\mu$ is the linear mass density.
We know that $T = Y A \frac{\Delta L}{L}$ and $\mu = \rho A$,where $A$ is the cross-sectional area.
Substituting these into the frequency formula:
$f = \frac{1}{2L} \sqrt{\frac{Y A \Delta L / L}{\rho A}} = \frac{1}{2L} \sqrt{\frac{Y \Delta L}{\rho L}}$.
Given values: $L = 0.5\,m$,$\Delta L = 3.2 \times 10^{-4}\,m$,$Y = 8 \times 10^{10}\,N/m^2$,$\rho = 8 \times 10^3\,kg/m^3$.
$f = \frac{1}{2 \times 0.5} \sqrt{\frac{8 \times 10^{10} \times 3.2 \times 10^{-4}}{8 \times 10^3 \times 0.5}}$.
$f = 1 \times \sqrt{\frac{25.6 \times 10^6}{4 \times 10^3}} = \sqrt{6.4 \times 10^3} = \sqrt{6400} = 80\,Hz$.
116
MediumMCQ
In an experiment with a sonometer,when a mass of $180\,g$ is attached to the string,it vibrates with a fundamental frequency of $30\,Hz$. When a mass $m$ is attached,the string vibrates with a fundamental frequency of $50\,Hz$. The value of $m$ is $.........\,g$.
A
$400$
B
$500$
C
$300$
D
$200$

Solution

(B) The fundamental frequency of a sonometer wire is given by $f = \frac{1}{2 \ell} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Since $T = Mg$ (where $M$ is the attached mass),we have $f \propto \sqrt{M}$.
Therefore,$\frac{f_2}{f_1} = \sqrt{\frac{M_2}{M_1}}$.
Given $f_1 = 30\,Hz$,$M_1 = 180\,g$,$f_2 = 50\,Hz$,and $M_2 = m$.
Substituting the values: $\frac{50}{30} = \sqrt{\frac{m}{180}}$.
Squaring both sides: $(\frac{5}{3})^2 = \frac{m}{180} \Rightarrow \frac{25}{9} = \frac{m}{180}$.
Solving for $m$: $m = \frac{25}{9} \times 180 = 25 \times 20 = 500\,g$.
117
DifficultMCQ
$A$ tuning fork resonates with a sonometer wire of length $1 \ m$ stretched with a tension of $6 \ N$. When the tension in the wire is changed to $54 \ N$,the same tuning fork produces $12$ beats per second with it. The frequency of the tuning fork is $Hz$.
A
$4$
B
$5$
C
$6$
D
$7$

Solution

(C) The frequency of a sonometer wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is length,$T$ is tension,and $\mu$ is linear mass density.
Since $L$ and $\mu$ are constant,$f \propto \sqrt{T}$.
Let $f_0$ be the frequency of the tuning fork.
For $T_1 = 6 \ N$,the wire resonates with the tuning fork,so $f_1 = f_0 = k\sqrt{6}$,where $k = \frac{1}{2L\sqrt{\mu}}$.
For $T_2 = 54 \ N$,the frequency of the wire is $f_2 = k\sqrt{54} = k\sqrt{9 \times 6} = 3k\sqrt{6} = 3f_0$.
The number of beats per second is $|f_2 - f_0| = 12$.
Substituting $f_2 = 3f_0$,we get $|3f_0 - f_0| = 12$.
$2f_0 = 12$,which gives $f_0 = 6 \ Hz$.
118
DifficultMCQ
$A$ sonometer wire of resonating length $90 \ cm$ has a fundamental frequency of $400 \ Hz$ when kept under some tension. The resonating length of the wire with a fundamental frequency of $600 \ Hz$ under the same tension is . . . . . . $cm$.
A
$30$
B
$40$
C
$50$
D
$60$

Solution

(D) For a sonometer wire,the fundamental frequency $f_0$ is given by $f_0 = \frac{v}{2L}$,where $v$ is the wave speed and $L$ is the resonating length.
Since the tension $T$ and mass per unit length $\mu$ are constant,the wave speed $v = \sqrt{\frac{T}{\mu}}$ remains constant.
Therefore,$f_0 L = \text{constant}$,which implies $f_1 L_1 = f_2 L_2$.
Given $f_1 = 400 \ Hz$,$L_1 = 90 \ cm$,and $f_2 = 600 \ Hz$.
Substituting the values: $400 \times 90 = 600 \times L_2$.
$L_2 = \frac{400 \times 90}{600} = \frac{36000}{600} = 60 \ cm$.
Solution diagram
119
MediumMCQ
$A$ string of length $1 \,m$ and mass $2 \times 10^{-5} \,kg$ is under tension $T$. When the string vibrates, two successive harmonics are found to occur at frequencies $750 \,Hz$ and $1000 \,Hz$. The value of tension $T$ is . . . . . . Newton.
A
$3$
B
$5$
C
$10$
D
$15$

Solution

(B) The frequency of the $p^{th}$ harmonic is given by $f_p = \frac{p}{2L} \sqrt{\frac{T}{\mu}}$, where $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density.
Given $L = 1 \,m$ and mass $m = 2 \times 10^{-5} \,kg$, the linear mass density $\mu = \frac{m}{L} = 2 \times 10^{-5} \,kg/m$.
Let the two successive harmonics be $p$ and $p+1$ with frequencies $f_p = 750 \,Hz$ and $f_{p+1} = 1000 \,Hz$.
Thus, $750 = \frac{p}{2 \times 1} \sqrt{\frac{T}{2 \times 10^{-5}}} \dots (1)$ and $1000 = \frac{p+1}{2 \times 1} \sqrt{\frac{T}{2 \times 10^{-5}}} \dots (2)$.
Dividing equation $(2)$ by $(1)$, we get $\frac{1000}{750} = \frac{p+1}{p} \Rightarrow \frac{4}{3} = \frac{p+1}{p} \Rightarrow 4p = 3p + 3 \Rightarrow p = 3$.
Substituting $p=3$ into equation $(1)$: $750 = \frac{3}{2} \sqrt{\frac{T}{2 \times 10^{-5}}} \Rightarrow 500 = \sqrt{\frac{T}{2 \times 10^{-5}}}$.
Squaring both sides: $250000 = \frac{T}{2 \times 10^{-5}} \Rightarrow T = 250000 \times 2 \times 10^{-5} = 5 \,N$.
120
Advanced
Answer the following by appropriately matching the lists based on the information given in the paragraph.
$A$ musical instrument is made using four different metal strings,$1, 2, 3$ and $4$ with mass per unit length $\mu, 2\mu, 3\mu$ and $4\mu$ respectively. The instrument is played by vibrating the strings by varying the free length in between the range $L_0$ and $2L_0$. It is found that in string-$1$ $(\mu)$ at free length $L_0$ and tension $T_0$ the fundamental mode frequency is $f_0$.
$List-I$ gives the above four strings while $List-II$ lists the magnitude of some quantity.
$List-I$$List-II$
$(I)$ String-$1$ $(\mu)$$(P) 1$
$(II)$ String-$2$ $(2\mu)$$(Q) 1/2$
$(III)$ String-$3$ $(3\mu)$$(R) 1/\sqrt{2}$
$(IV)$ String-$4$ $(4\mu)$$(S) 1/\sqrt{3}$
$(T) 3/16$
$(U) 1/16$

$(1)$ If the tension in each string is $T_0$,the correct match for the fundamental frequency in $f_0$ units will be,
$(1)$ $I \rightarrow P, II \rightarrow R, III \rightarrow S, IV \rightarrow Q$
$(2)$ $I \rightarrow P, II \rightarrow Q, III \rightarrow T, IV \rightarrow S$
$(3)$ $I \rightarrow Q, II \rightarrow S, III \rightarrow R, IV \rightarrow P$
$(4)$ $I \rightarrow Q, II \rightarrow P, III \rightarrow R, IV \rightarrow T$
$(2)$ The lengths of the strings $1, 2, 3$ and $4$ are kept fixed at $L_0, 3L_0/2, 5L_0/4$ and $7L_0/4$,respectively. Strings $1, 2, 3$ and $4$ are vibrated at their $1^{st}, 3^{rd}, 5^{th}$ and $14^{th}$ harmonics,respectively,such that all the strings have the same frequency. The correct match for the tension in the four strings in the units of $T_0$ will be.
$(1)$ $I \rightarrow P, II \rightarrow Q, III \rightarrow T, IV \rightarrow U$
$(2)$ $I \rightarrow T, II \rightarrow Q, III \rightarrow R, IV \rightarrow U$
$(3)$ $I \rightarrow P, II \rightarrow Q, III \rightarrow R, IV \rightarrow T$
$(4)$ $I \rightarrow P, II \rightarrow R, III \rightarrow T, IV \rightarrow U$

Solution

(A) For the fundamental mode,the frequency is given by $f = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$.
$(1)$ With $T = T_0$ and $L = L_0$,$f_n = \frac{n}{2L_0} \sqrt{\frac{T_0}{\mu_n}}$.
For string $1$: $f_1 = \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu}} = f_0 \rightarrow (P)$.
For string $2$: $f_2 = \frac{1}{2L_0} \sqrt{\frac{T_0}{2\mu}} = \frac{f_0}{\sqrt{2}} \rightarrow (R)$.
For string $3$: $f_3 = \frac{1}{2L_0} \sqrt{\frac{T_0}{3\mu}} = \frac{f_0}{\sqrt{3}} \rightarrow (S)$.
For string $4$: $f_4 = \frac{1}{2L_0} \sqrt{\frac{T_0}{4\mu}} = \frac{f_0}{2} \rightarrow (Q)$.
Thus,$I \rightarrow P, II \rightarrow R, III \rightarrow S, IV \rightarrow Q$.
$(2)$ Given $f = f_0$ for all strings.
For string $1$: $f_0 = \frac{1}{2L_0} \sqrt{\frac{T_1}{\mu}} = f_0 \Rightarrow T_1 = T_0 \rightarrow (P)$.
For string $2$: $f_0 = \frac{3}{2(3L_0/2)} \sqrt{\frac{T_2}{2\mu}} = \frac{1}{L_0} \sqrt{\frac{T_2}{2\mu}} = f_0 \Rightarrow T_2 = T_0/2 \rightarrow (Q)$.
For string $3$: $f_0 = \frac{5}{2(5L_0/4)} \sqrt{\frac{T_3}{3\mu}} = \frac{2}{L_0} \sqrt{\frac{T_3}{3\mu}} = f_0 \Rightarrow T_3 = 3T_0/16 \rightarrow (T)$.
For string $4$: $f_0 = \frac{14}{2(7L_0/4)} \sqrt{\frac{T_4}{4\mu}} = \frac{4}{L_0} \sqrt{\frac{T_4}{4\mu}} = f_0 \Rightarrow T_4 = T_0/16 \rightarrow (U)$.
Thus,$I \rightarrow P, II \rightarrow Q, III \rightarrow T, IV \rightarrow U$.
Solution diagram
121
DifficultMCQ
$A$ horizontal stretched string,fixed at two ends,is vibrating in its fifth harmonic according to the equation,$y(x, t) = (0.01 \ m) \sin[(62.8 \ m^{-1}) x] \cos[(628 \ s^{-1}) t]$. Assuming $\pi = 3.14$,the correct statement$(s)$ is (are) :
$(A)$ The number of nodes is $5$.
$(B)$ The length of the string is $0.25 \ m$.
$(C)$ The maximum displacement of the midpoint of the string from its equilibrium position is $0.01 \ m$.
$(D)$ The fundamental frequency is $100 \ Hz$.
A
$(B, D)$
B
$(B, C)$
C
$(A, D)$
D
$(C, D)$

Solution

(B) The given equation is $y(x, t) = (0.01 \ m) \sin(62.8x) \cos(628t)$.
$(A)$ For the $n^{th}$ harmonic,the number of loops is $n = 5$. The number of nodes is $n + 1 = 5 + 1 = 6$. Thus,statement $(A)$ is incorrect.
$(B)$ The wave number $k = 62.8 \ m^{-1}$. Since $k = \frac{2\pi}{\lambda}$,we have $\lambda = \frac{2 \times 3.14}{62.8} = 0.1 \ m$. For the $5^{th}$ harmonic,the length of the string $L = \frac{5\lambda}{2} = \frac{5 \times 0.1}{2} = 0.25 \ m$. Thus,statement $(B)$ is correct.
$(C)$ The amplitude of the standing wave is $A(x) = 0.01 \sin(62.8x)$. At the midpoint $x = \frac{L}{2} = 0.125 \ m$,$A(0.125) = 0.01 \sin(62.8 \times 0.125) = 0.01 \sin(7.85) \approx 0.01 \sin(2.5\pi) = 0.01 \times 1 = 0.01 \ m$. Thus,statement $(C)$ is correct.
$(D)$ The angular frequency $\omega = 628 \ rad/s$. The frequency $f = \frac{\omega}{2\pi} = \frac{628}{2 \times 3.14} = 100 \ Hz$. The fundamental frequency $f_1 = \frac{f}{n} = \frac{100}{5} = 20 \ Hz$. Thus,statement $(D)$ is incorrect.
Therefore,the correct statements are $(B)$ and $(C)$.
Solution diagram
122
MediumMCQ
$A$ wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of $25 \ Hz$. The mass of the wire is $2 \ g$ and its linear mass density is $4 \times 10^{-3} \ kg/m$. What is the tension in the string (in $N$)?
A
$5$
B
$10$
C
$2.5$
D
$20$

Solution

(C) The frequency of the fundamental mode of a stretched string is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.
Given: $n = 25 \ Hz$,$m = 2 \ g = 2 \times 10^{-3} \ kg$,and $\mu = 4 \times 10^{-3} \ kg/m$.
First,calculate the length $L$ of the wire using the relation $\mu = \frac{m}{L}$:
$L = \frac{m}{\mu} = \frac{2 \times 10^{-3} \ kg}{4 \times 10^{-3} \ kg/m} = 0.5 \ m$.
Now,substitute the values into the frequency formula:
$25 = \frac{1}{2 \times 0.5} \sqrt{\frac{T}{4 \times 10^{-3}}}$
$25 = \frac{1}{1} \sqrt{\frac{T}{4 \times 10^{-3}}}$
Squaring both sides:
$625 = \frac{T}{4 \times 10^{-3}}$
$T = 625 \times 4 \times 10^{-3} = 2500 \times 10^{-3} = 2.5 \ N$.
123
MediumMCQ
Two vibrating strings $A$ and $B$ of the same material but lengths $3 L$ and $2 L$ have radii $2 r$ and $3 r$ respectively. They are stretched under the same tension. String $A$ vibrates in the second overtone and string $B$ in the fundamental mode. The ratio of their frequencies $n_A / n_B$ will be:
A
$2: 1$
B
$3: 1$
C
$4: 1$
D
$3: 2$

Solution

(B) The frequency of a vibrating string is given by $n = \frac{p}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu = \rho \pi r^2$ is the linear mass density.
Since the material is the same,density $\rho$ is constant. Given tension $T$ is also constant,the frequency is $n \propto \frac{p}{Lr}$.
For string $A$: Length $L_A = 3L$,radius $r_A = 2r$,and it vibrates in the second overtone,so $p_A = 3$.
Thus,$n_A \propto \frac{3}{(3L)(2r)} = \frac{1}{2Lr}$.
For string $B$: Length $L_B = 2L$,radius $r_B = 3r$,and it vibrates in the fundamental mode,so $p_B = 1$.
Thus,$n_B \propto \frac{1}{(2L)(3r)} = \frac{1}{6Lr}$.
The ratio is $\frac{n_A}{n_B} = \frac{1/2Lr}{1/6Lr} = \frac{6}{2} = 3$.
Therefore,the ratio $n_A / n_B = 3: 1$.
124
DifficultMCQ
Length of a sonometer wire is either $95 \ cm$ or $100 \ cm$. In both the cases,a tuning fork produces $5 \ beats/sec$ with the wire. The frequency of the tuning fork is- (in $Hz$)
A
$192$
B
$195$
C
$190$
D
$200$

Solution

(B) The frequency of a sonometer wire is given by $f_w = \frac{v}{2L}$.
Since the tuning fork produces $5 \ beats/sec$ in both cases,the frequency of the wire $f_w$ is either $(f + 5)$ or $(f - 5)$.
For the shorter length $(95 \ cm)$,the frequency is higher: $f_1 = \frac{v}{2 \times 0.95} = f + 5 \dots (1)$
For the longer length $(100 \ cm)$,the frequency is lower: $f_2 = \frac{v}{2 \times 1.00} = f - 5 \dots (2)$
Dividing $(1)$ by $(2)$:
$\frac{f+5}{f-5} = \frac{1.00}{0.95} = \frac{100}{95} = \frac{20}{19}$
$19(f + 5) = 20(f - 5)$
$19f + 95 = 20f - 100$
$f = 195 \ Hz$.
125
MediumMCQ
$A$ tuning fork produces $4$ beats per second when sounded with a sonometer wire of vibrating length $48 \ cm$. It also produces $4$ beats per second when the vibrating length is $50 \ cm$,keeping the tension in the wire the same. What is the frequency of the tuning fork in $Hz$?
A
$196$
B
$284$
C
$375$
D
$460$

Solution

(A) The frequency of a sonometer wire is given by $f = \frac{v}{2L}$,where $v$ is constant as tension is constant. Thus,$f \propto \frac{1}{L}$,which means $fL = \text{constant}$.
Let $f$ be the frequency of the tuning fork.
For length $L_1 = 48 \ cm$,the wire frequency $f_1 = \frac{k}{48}$. Since it produces $4$ beats,$f_1 = f + 4$ or $f_1 = f - 4$. Since $L_1 < L_2$,$f_1 > f_2$,so $f_1 = f + 4$.
For length $L_2 = 50 \ cm$,the wire frequency $f_2 = \frac{k}{50}$. Since it produces $4$ beats,$f_2 = f - 4$.
Thus,$f_1 L_1 = f_2 L_2 \implies (f + 4) \times 48 = (f - 4) \times 50$.
$48f + 192 = 50f - 200$.
$2f = 392$.
$f = 196 \ Hz$.
Solution diagram
126
DifficultMCQ
$A$ steel wire of length $1 \ m$ and mass $0.1 \ kg$ and having a uniform cross-sectional area of $10^{-6} \ m^2$ is rigidly fixed at both ends. The temperature of the wire is lowered by $20^{\circ} C$. If the wire is vibrating in its fundamental mode,find the frequency (in $Hz$).
$(Y_{\text{steel}} = 2 \times 10^{11} \ N/m^2, \alpha_{\text{steel}} = 1.21 \times 10^{-5} /^{\circ} C)$
A
$11$
B
$20$
C
$15$
D
$10$

Solution

(A) The thermal strain produced due to a change in temperature $\Delta t$ is given by $\text{Strain} = \alpha \Delta t$.
Using Hooke's Law,the thermal stress is $\text{Stress} = Y \times \text{Strain} = Y \alpha \Delta t$.
The tension $T$ in the wire is $T = \text{Area} \times \text{Stress} = A Y \alpha \Delta t$.
Given $A = 10^{-6} \ m^2$,$Y = 2 \times 10^{11} \ N/m^2$,$\alpha = 1.21 \times 10^{-5} /^{\circ} C$,and $\Delta t = 20^{\circ} C$,we have:
$T = 10^{-6} \times 2 \times 10^{11} \times 1.21 \times 10^{-5} \times 20 = 48.4 \ N$.
The linear mass density $\mu = \frac{m}{\ell} = \frac{0.1 \ kg}{1 \ m} = 0.1 \ kg/m$.
The fundamental frequency $f$ is given by $f = \frac{1}{2 \ell} \sqrt{\frac{T}{\mu}}$.
Substituting the values: $f = \frac{1}{2 \times 1} \sqrt{\frac{48.4}{0.1}} = \frac{1}{2} \sqrt{484} = \frac{22}{2} = 11 \ Hz$.
127
MediumMCQ
$A$ violin emits sound waves of frequency $n_1$ under tension $T$. When tension is increased by $44\%$,keeping the length and mass per unit length constant,the frequency of sound waves becomes $n_2$. The ratio of frequency $n_2$ to frequency $n_1$ is:
A
$5: 6$
B
$6: 7$
C
$6: 5$
D
$7: 6$

Solution

(C) The frequency of a stretched string is given by the formula $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length,$T$ is the tension,and $\mu$ is the mass per unit length.
Initially,the frequency is $n_1 = \frac{1}{2L} \sqrt{\frac{T_1}{\mu}}$.
When the tension is increased by $44\%$,the new tension $T_2$ becomes $T_2 = T_1 + 0.44T_1 = 1.44T_1$.
The new frequency $n_2$ is given by $n_2 = \frac{1}{2L} \sqrt{\frac{1.44T_1}{\mu}}$.
Taking the ratio of $n_2$ to $n_1$:
$\frac{n_2}{n_1} = \frac{\frac{1}{2L} \sqrt{\frac{1.44T_1}{\mu}}}{\frac{1}{2L} \sqrt{\frac{T_1}{\mu}}} = \sqrt{\frac{1.44T_1}{T_1}} = \sqrt{1.44} = 1.2$.
Converting $1.2$ to a fraction,we get $1.2 = \frac{12}{10} = \frac{6}{5}$.
Thus,the ratio $n_2:n_1$ is $6:5$.
128
MediumMCQ
When a string of length $l$ is divided into three segments of length $l_1, l_2$ and $l_3$,the fundamental frequencies of the three segments are $n_1, n_2$ and $n_3$ respectively. The original fundamental frequency $n$ of the string is:
A
$n = n_1 + n_2 + n_3$
B
$\sqrt{n} = \sqrt{n_1} + \sqrt{n_2} + \sqrt{n_3}$
C
$\frac{1}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3}$
D
$\frac{1}{\sqrt{n}} = \frac{1}{\sqrt{n_1}} + \frac{1}{\sqrt{n_2}} + \frac{1}{\sqrt{n_3}}$

Solution

(C) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is tension and $\mu$ is mass per unit length.
Since $T$ and $\mu$ are constant for the segments,we have $n \propto \frac{1}{l}$,which implies $nl = k$ (constant).
Given that the total length $l = l_1 + l_2 + l_3$,we can express each segment length as $l_1 = \frac{k}{n_1}$,$l_2 = \frac{k}{n_2}$,and $l_3 = \frac{k}{n_3}$.
The original length is $l = \frac{k}{n}$.
Substituting these into the length equation: $\frac{k}{n} = \frac{k}{n_1} + \frac{k}{n_2} + \frac{k}{n_3}$.
Dividing both sides by $k$,we get $\frac{1}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3}$.
129
MediumMCQ
$A$ stretched wire of length $260 \ cm$ is set into vibrations. It is divided into three segments whose frequencies are in the ratio $2:3:4$. Their lengths must be
A
$80 \ cm, 60 \ cm, 120 \ cm$
B
$120 \ cm, 80 \ cm, 60 \ cm$
C
$60 \ cm, 80 \ cm, 120 \ cm$
D
$120 \ cm, 60 \ cm, 80 \ cm$

Solution

(B) The frequency produced by a stretched wire is given by $f = \frac{p}{2l} \sqrt{\frac{T}{m}}$.
Here,$p$ is the number of loops,$l$ is the length,$T$ is the tension,and $m$ is the mass per unit length.
Since $T$ and $m$ are constant for the same wire,we have $f \propto \frac{1}{l}$.
Given the ratio of frequencies $f_1 : f_2 : f_3 = 2 : 3 : 4$,the ratio of lengths is $l_1 : l_2 : l_3 = \frac{1}{2} : \frac{1}{3} : \frac{1}{4}$.
Multiplying by the least common multiple $(12)$,we get $l_1 : l_2 : l_3 = 6 : 4 : 3$.
The sum of the ratio parts is $6 + 4 + 3 = 13$.
Given the total length $L = 260 \ cm$,the individual lengths are:
$l_1 = \frac{6}{13} \times 260 = 120 \ cm$
$l_2 = \frac{4}{13} \times 260 = 80 \ cm$
$l_3 = \frac{3}{13} \times 260 = 60 \ cm$.
130
MediumMCQ
$A$ sonometer wire $49 \ cm$ long is in unison with a tuning fork of frequency '$n$'. If the length of the wire is decreased by $1 \ cm$ and it is vibrated with the same tuning fork,$6$ beats are heard per second. The value of '$n$' is (in $Hz$)
A
$256$
B
$288$
C
$320$
D
$384$

Solution

(B) For a sonometer wire,the frequency $n$ is inversely proportional to the length $L$ $(n \propto 1/L)$,so $n_1 L_1 = n_2 L_2$.
Given $L_1 = 49 \ cm$ and $L_2 = 48 \ cm$,and the initial frequency $n_1 = n$.
Then $n \times 49 = n_2 \times 48$,which gives $n_2 = \frac{49}{48} n$.
Since the length is decreased,the frequency increases,so $n_2 > n_1$.
The beat frequency is given by $n_2 - n_1 = 6$.
Substituting the value of $n_2$: $\frac{49}{48} n - n = 6$.
$\frac{n}{48} = 6$.
$n = 6 \times 48 = 288 \ Hz$.
131
DifficultMCQ
At the poles of the earth,a stretched wire of a given length vibrates in unison with a tuning fork. At the equator of the earth,for the same setting,to produce resonance with the same fork,the vibrating length of the wire
A
should be decreased.
B
should be increased.
C
should be same.
D
should be three times the original.

Solution

(A) The frequency of vibration of a stretched wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$.
Here,$T$ is the tension in the wire,which is provided by the weight of a mass $M$ suspended from it,so $T = Mg$.
Thus,$n = \frac{1}{2l} \sqrt{\frac{Mg}{m}}$.
For the frequency $n$ to remain constant with the same tuning fork,we have $\frac{1}{l} \sqrt{g} = \text{constant}$.
This implies $l \propto \sqrt{g}$.
Since the acceleration due to gravity $g$ is greater at the poles $(g_p)$ than at the equator $(g_e)$,i.e.,$g_p > g_e$,it follows that the length $l$ must be adjusted.
To maintain the same frequency at the equator where $g$ is smaller,the length $l$ must be decreased to compensate for the decrease in $g$ such that the ratio $\frac{\sqrt{g}}{l}$ remains constant.
Therefore,the vibrating length of the wire should be decreased.
132
EasyMCQ
At the poles,a stretched wire of a given length vibrates in unison with a tuning fork. At the equator,for the same setting to produce resonance with the same fork,the vibrating length of the wire
A
should be increased.
B
should be $3$ times the original length.
C
should be same.
D
should be decreased.

Solution

(D) The frequency of vibration of a stretched wire is given by $n = \frac{1}{2\ell} \sqrt{\frac{T}{m}}$.
Assuming the tension $T$ is provided by a mass $M$ hanging from the wire,$T = Mg$,so $n = \frac{1}{2\ell} \sqrt{\frac{Mg}{m}}$.
Let $\ell_1$ and $g_1$ be the length and acceleration due to gravity at the poles,and $\ell_2$ and $g_2$ be the length and acceleration due to gravity at the equator.
Since the frequency $n$ remains the same for the same tuning fork,we have $\frac{1}{2\ell_1} \sqrt{\frac{Mg_1}{m}} = \frac{1}{2\ell_2} \sqrt{\frac{Mg_2}{m}}$.
This simplifies to $\frac{\sqrt{g_1}}{\ell_1} = \frac{\sqrt{g_2}}{\ell_2}$,which implies $\frac{\ell_2}{\ell_1} = \sqrt{\frac{g_2}{g_1}}$.
Since the acceleration due to gravity is greater at the poles than at the equator $(g_1 > g_2)$,it follows that $\ell_1 > \ell_2$.
Therefore,to maintain resonance at the equator,the vibrating length of the wire must be decreased.
133
EasyMCQ
In a sonometer experiment,a string of length $L$ under tension vibrates in its second overtone between two bridges. The amplitude of vibration is maximum at:
A
$\frac{L}{6}, \frac{L}{2}, \frac{5L}{6}$
B
$\frac{L}{8}, \frac{L}{4}, \frac{L}{2}$
C
$\frac{L}{2}, \frac{L}{4}, \frac{L}{6}$
D
$\frac{L}{3}, \frac{2L}{3}, \frac{5L}{6}$

Solution

(A) In the second overtone of a string fixed at both ends,the string vibrates in $3$ loops (segments).
For a string of length $L$ vibrating in $n$ loops,the positions of the antinodes (where amplitude is maximum) are given by $x = \frac{(2k-1)L}{2n}$,where $k = 1, 2, ..., n$.
Here,$n = 3$ (second overtone corresponds to the $3^{rd}$ harmonic).
For $k = 1$: $x_1 = \frac{(2(1)-1)L}{2(3)} = \frac{L}{6}$.
For $k = 2$: $x_2 = \frac{(2(2)-1)L}{2(3)} = \frac{3L}{6} = \frac{L}{2}$.
For $k = 3$: $x_3 = \frac{(2(3)-1)L}{2(3)} = \frac{5L}{6}$.
Thus,the amplitude is maximum at $\frac{L}{6}, \frac{L}{2}, \text{ and } \frac{5L}{6}$.
Solution diagram
134
EasyMCQ
$A$ stationary wave is produced along a stretched string of length $80 \,cm$. The resonant frequencies of the string are $90 \,Hz$, $150 \,Hz$, and $210 \,Hz$. The speed of the transverse wave in the string is: (in $\,m/s$)
A
$45$
B
$75$
C
$48$
D
$80$

Solution

(C) The resonant frequencies are given as $f_1 = 90 \,Hz$, $f_2 = 150 \,Hz$, and $f_3 = 210 \,Hz$.
The difference between consecutive resonant frequencies is $\Delta f = 150 - 90 = 60 \,Hz$ and $210 - 150 = 60 \,Hz$.
For a string fixed at both ends, the resonant frequencies are integer multiples of the fundamental frequency $f_0$. Thus, $\Delta f = f_0 = 60 \,Hz$.
The given frequencies are $f_1 = 1.5 f_0$ (not possible for both ends fixed) or they are harmonics of a string fixed at one end. However, if we assume the fundamental frequency $f_0 = 30 \,Hz$ (since $90 = 3 \times 30$, $150 = 5 \times 30$, $210 = 7 \times 30$), these are odd harmonics of a string fixed at both ends where the fundamental is $30 \,Hz$.
Using $f_n = \frac{n v}{2L}$, for $n=3$, $f_3 = \frac{3v}{2L} = 90 \,Hz$.
Given $L = 0.8 \,m$, we have $90 = \frac{3v}{2 \times 0.8}$.
$90 = \frac{3v}{1.6} \implies 3v = 144 \implies v = 48 \,m/s$.
135
MediumMCQ
In Melde's experiment,when the tension decreases by $0.009 \ kg-wt$,the number of loops changes from $4$ to $5$. The initial tension is
A
$0.036 \ kg-wt$.
B
$0.009 \ kg-wt$.
C
$0.018 \ kg-wt$.
D
$0.025 \ kg-wt$.

Solution

(D) In Melde's experiment,the frequency of the vibrating string is given by $f = \frac{P}{2L} \sqrt{\frac{T}{\mu}}$,where $P$ is the number of loops,$T$ is the tension,$L$ is the length,and $\mu$ is the linear mass density.
Since the frequency $f$ and the length $L$ remain constant,we have $P \propto \frac{1}{\sqrt{T}}$,which implies $T P^{2} = \text{constant}$.
Let the initial tension be $T_1$ and the final tension be $T_2 = T_1 - 0.009 \ kg-wt$.
Given $P_1 = 4$ and $P_2 = 5$.
Using the relation $T_1 P_1^{2} = T_2 P_2^{2}$:
$T_1 (4)^{2} = (T_1 - 0.009) (5)^{2}$
$16 T_1 = 25 T_1 - 0.009 \times 25$
$25 T_1 - 16 T_1 = 0.225$
$9 T_1 = 0.225$
$T_1 = \frac{0.225}{9} = 0.025 \ kg-wt$.
136
EasyMCQ
If you set up the $7^{th}$ overtone on a string fixed at both ends,how many nodes and antinodes are set up in it?
A
$9, 8$
B
$8, 9$
C
$7, 8$
D
$8, 7$

Solution

(A) For a string fixed at both ends,the frequency of the $n^{th}$ harmonic is given by $f_n = n f_1$,where $n = 1, 2, 3, \dots$ is the mode number.
The $n^{th}$ harmonic corresponds to $(n-1)^{th}$ overtone.
Given that we have the $7^{th}$ overtone,we have $n-1 = 7$,which implies $n = 8$.
Thus,the $7^{th}$ overtone is the $8^{th}$ harmonic.
In the $n^{th}$ harmonic,the number of loops is $n$.
Therefore,for the $8^{th}$ harmonic,there are $8$ loops.
For a string with $n$ loops,the number of nodes is $n+1$ and the number of antinodes is $n$.
For $n = 8$,the number of nodes is $8 + 1 = 9$ and the number of antinodes is $8$.
Solution diagram
137
DifficultMCQ
$A$ uniform wire of length $L$,diameter $D$,and density $\rho$ is stretched by a tension $T$. The frequency $f$ of the wire is proportional to:
A
$f \propto \frac{L}{D}$
B
$f \propto \frac{1}{L D}$
C
$f \propto \frac{1}{L \sqrt{D}}$
D
$f \propto \frac{1}{L D^2}$

Solution

(B) The fundamental frequency $f$ of a stretched wire is given by the formula: $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Linear mass density $\mu$ is defined as mass per unit length,which can be expressed in terms of density $\rho$ and cross-sectional area $A$ as $\mu = A \rho$.
Since the wire has a circular cross-section with diameter $D$,the area $A = \pi \frac{D^2}{4}$.
Substituting this into the expression for $\mu$,we get $\mu = \frac{\pi D^2 \rho}{4}$.
Now,substitute $\mu$ back into the frequency formula: $f = \frac{1}{2L} \sqrt{\frac{T}{\frac{\pi D^2 \rho}{4}}} = \frac{1}{2L} \sqrt{\frac{4T}{\pi D^2 \rho}} = \frac{1}{2L} \cdot \frac{2}{D} \sqrt{\frac{T}{\pi \rho}} = \frac{1}{LD} \sqrt{\frac{T}{\pi \rho}}$.
From this expression,it is clear that $f \propto \frac{1}{LD}$.
138
MediumMCQ
$A$ wire of length $L$,diameter $d$,and density of material $\rho$ is under tension $T$,having a fundamental frequency of vibration $n_A$. Another wire of length $2L$,tension $2T$,density $2\rho$,and diameter $3d$ has a fundamental frequency of vibration $n_B$. The ratio $n_B : n_A$ is
A
$1 : 2$
B
$1 : 4$
C
$1 : 6$
D
$1 : 8$

Solution

(C) The fundamental frequency of a stretched wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu$ is the mass per unit length.
Since $\mu = \text{Area} \times \text{Density} = (\pi r^2) \rho = \pi (d/2)^2 \rho = \frac{\pi d^2 \rho}{4}$,we can write $n = \frac{1}{2L} \sqrt{\frac{T}{\frac{\pi d^2 \rho}{4}}} = \frac{1}{Ld} \sqrt{\frac{T}{\pi \rho}}$.
For wire $A$: $n_A = \frac{1}{Ld} \sqrt{\frac{T}{\pi \rho}}$.
For wire $B$: $n_B = \frac{1}{(2L)(3d)} \sqrt{\frac{2T}{\pi (2\rho)}} = \frac{1}{6Ld} \sqrt{\frac{T}{\pi \rho}}$.
Taking the ratio $n_B : n_A = \frac{\frac{1}{6Ld} \sqrt{\frac{T}{\pi \rho}}}{\frac{1}{Ld} \sqrt{\frac{T}{\pi \rho}}} = \frac{1}{6}$.
Thus,the ratio $n_B : n_A$ is $1 : 6$.
139
MediumMCQ
Two uniform wires of same material are vibrating under the same tension. If the $1^{\text{st}}$ overtone of the $1^{\text{st}}$ wire is equal to the $2^{\text{nd}}$ overtone of the $2^{\text{nd}}$ wire and the radius of the $1^{\text{st}}$ wire is twice the radius of the $2^{\text{nd}}$ wire,find the ratio of the length of the $1^{\text{st}}$ wire to that of the $2^{\text{nd}}$ wire.
A
$1: 3$
B
$3: 1$
C
$2: 3$
D
$3: 2$

Solution

(A) The frequency of the $n^{\text{th}}$ harmonic for a stretched string is given by $f = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$.
Thus,$f = \frac{n}{2L r} \sqrt{\frac{T}{\pi \rho}}$.
For the $1^{\text{st}}$ wire,the $1^{\text{st}}$ overtone is the $2^{\text{nd}}$ harmonic $(n_1 = 2)$.
So,$f_1 = \frac{2}{2L_1 r_1} \sqrt{\frac{T}{\pi \rho}} = \frac{1}{L_1 r_1} \sqrt{\frac{T}{\pi \rho}}$.
For the $2^{\text{nd}}$ wire,the $2^{\text{nd}}$ overtone is the $3^{\text{rd}}$ harmonic $(n_2 = 3)$.
So,$f_2 = \frac{3}{2L_2 r_2} \sqrt{\frac{T}{\pi \rho}}$.
Given $f_1 = f_2$ and $r_1 = 2r_2$:
$\frac{1}{L_1 (2r_2)} = \frac{3}{2L_2 r_2}$.
$\frac{1}{2L_1} = \frac{3}{2L_2} \implies \frac{L_1}{L_2} = \frac{1}{3}$.
Therefore,the ratio is $1: 3$.
140
EasyMCQ
The fundamental frequency of a sonometer wire is $50 \text{ Hz}$ for a given length and tension. If the length is increased by $25 \%$ while keeping the tension constant,what is the percentage change in the frequency of the second harmonic?
A
increased by $20 \%$
B
decreased by $20 \%$
C
increased by $10 \%$
D
decreased by $10 \%$

Solution

(B) The fundamental frequency of a sonometer wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.
Since $T$ and $\mu$ are constant,$f \propto \frac{1}{L}$.
Let the initial length be $L_1 = L$ and the initial fundamental frequency be $f_1 = 50 \text{ Hz}$.
The new length is $L_2 = L + 0.25L = 1.25L = \frac{5}{4}L$.
The new fundamental frequency $f_2$ is given by $\frac{f_2}{f_1} = \frac{L_1}{L_2} = \frac{L}{1.25L} = \frac{1}{1.25} = 0.8$.
So,$f_2 = 0.8 \times 50 \text{ Hz} = 40 \text{ Hz}$.
The frequency of the second harmonic is $f'_n = n \times f_n$. For the second harmonic $(n=2)$,$f'_2 = 2 \times f_2 = 2 \times 40 \text{ Hz} = 80 \text{ Hz}$.
The initial frequency of the second harmonic was $f'_1 = 2 \times f_1 = 2 \times 50 \text{ Hz} = 100 \text{ Hz}$.
The change in frequency is $\Delta f = f'_2 - f'_1 = 80 - 100 = -20 \text{ Hz}$.
The percentage change is $\frac{\Delta f}{f'_1} \times 100 = \frac{-20}{100} \times 100 = -20 \%$.
Thus,the frequency of the second harmonic decreases by $20 \%$.
141
MediumMCQ
Two tuning forks when sounded together produce $4$ beats per second. One of the forks is in unison with $23 \ cm$ length of a sonometer wire and the other with $24 \ cm$ length of the same wire. The frequencies of the two tuning forks are
A
$96 \ Hz, 92 \ Hz$
B
$92 \ Hz, 88 \ Hz$
C
$72 \ Hz, 68 \ Hz$
D
$48 \ Hz, 44 \ Hz$

Solution

(A) For a sonometer wire,the frequency $f$ is inversely proportional to the length $l$ of the wire,i.e.,$f \propto 1/l$ or $f \cdot l = k$ (constant).
Let the frequencies of the two tuning forks be $f_1$ and $f_2$.
Given $f_1 \cdot 23 = f_2 \cdot 24 = k$.
This implies $f_1 = k/23$ and $f_2 = k/24$.
Since $f_1 > f_2$,the beat frequency is $f_1 - f_2 = 4$.
Substituting the values: $k/23 - k/24 = 4$.
Taking $k$ as common: $k(24 - 23) / (23 \cdot 24) = 4$.
$k / 552 = 4 \implies k = 2208$.
Now,$f_1 = 2208 / 23 = 96 \ Hz$.
And $f_2 = 2208 / 24 = 92 \ Hz$.
Thus,the frequencies are $96 \ Hz$ and $92 \ Hz$.
142
MediumMCQ
The fundamental frequency of a sonometer wire is $n$. If the tension is increased $3$ times,the length is increased $3$ times,and the diameter is increased $2$ times,what will be the new frequency?
A
$\sqrt{\frac{3}{2}} n$
B
$\frac{\sqrt{3}}{2} n$
C
$\frac{n}{2 \sqrt{3}}$
D
$2 \sqrt{3} n$

Solution

(C) The fundamental frequency $n$ of a sonometer wire is given by the formula: $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length,$T$ 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),the frequency can be written as: $n = \frac{1}{2L} \sqrt{\frac{T}{\pi r^2 \rho}} = \frac{1}{2Lr} \sqrt{\frac{T}{\pi \rho}}$.
Given that the diameter $D = 2r$,we have $n \propto \frac{1}{LD} \sqrt{T}$.
Let the initial frequency be $n_1 = n$. The new frequency $n_2$ is given by the ratios of the changes:
$n_2 = n_1 \times \left( \frac{L_1}{L_2} \right) \times \left( \frac{D_1}{D_2} \right) \times \sqrt{\frac{T_2}{T_1}}$.
Given $L_2 = 3L_1$,$D_2 = 2D_1$,and $T_2 = 3T_1$:
$n_2 = n \times \left( \frac{1}{3} \right) \times \left( \frac{1}{2} \right) \times \sqrt{3} = n \times \frac{\sqrt{3}}{6} = \frac{\sqrt{3}}{2 \times 3} n = \frac{\sqrt{3}}{2 \times \sqrt{3} \times \sqrt{3}} n = \frac{n}{2 \sqrt{3}}$.
Thus,the new frequency is $\frac{n}{2 \sqrt{3}}$.
143
MediumMCQ
At poles,a stretched wire of a given length vibrates in unison with a tuning fork. At the equator,for the same setting,to produce resonance with the same fork,the vibrating length of the wire
A
should be decreased.
B
should be increased.
C
should be same.
D
should be three times.

Solution

(A) The frequency of a vibrating stretched wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the vibrating length,$T$ is the tension,and $\mu$ is the mass per unit length.
Since the frequency $f$ of the tuning fork is constant,we have $L \propto \sqrt{T}$.
The tension $T$ in the wire is provided by a suspended mass $M$,so $T = Mg$,where $g$ is the acceleration due to gravity.
Thus,$L \propto \sqrt{g}$.
At the equator,the value of $g$ is less than at the poles $(g_{equator} < g_{poles})$.
Since $L$ is directly proportional to $\sqrt{g}$,the vibrating length $L$ must be decreased at the equator to maintain the same frequency $f$.
144
MediumMCQ
$A$ tuning fork gives $5$ beats per second with $40 \ cm$ length of a sonometer wire. If the length of the wire is shortened by $1 \ cm$,the number of beats is still the same. The frequency of the fork is (in $Hz$)
A
$390$
B
$395$
C
$400$
D
$405$

Solution

(B) The frequency of a sonometer wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,which implies $f \propto \frac{1}{L}$.
Let the frequency of the tuning fork be $n$.
For length $L_1 = 40 \ cm$,the frequency of the wire is $f_1 = \frac{k}{40}$. Since it produces $5$ beats,$f_1 = n \pm 5$.
For length $L_2 = 39 \ cm$,the frequency of the wire is $f_2 = \frac{k}{39}$. Since it produces $5$ beats,$f_2 = n \pm 5$.
Since $L_2 < L_1$,$f_2 > f_1$. Thus,$f_1 = n - 5$ and $f_2 = n + 5$.
So,$\frac{k}{40} = n - 5$ and $\frac{k}{39} = n + 5$.
From the first equation,$k = 40(n - 5)$.
From the second equation,$k = 39(n + 5)$.
Equating the two: $40n - 200 = 39n + 195$.
$40n - 39n = 195 + 200$.
$n = 395 \ Hz$.
145
EasyMCQ
In a Sonometer experiment,the frequency of the tuning fork used is $288 \ Hz$. Harmonics will '$NOT$' be produced at the frequency: (in $Hz$)
A
$288$
B
$576$
C
$844$
D
$864$

Solution

(C) In a Sonometer experiment,the string vibrates in harmonics which are integer multiples of the fundamental frequency $(f_0)$.
Given the fundamental frequency $f_0 = 288 \ Hz$.
The frequencies of the harmonics are given by $f_n = n \times f_0$,where $n = 1, 2, 3, ...$.
Calculating the first few harmonics:
$n=1: f_1 = 1 \times 288 = 288 \ Hz$
$n=2: f_2 = 2 \times 288 = 576 \ Hz$
$n=3: f_3 = 3 \times 288 = 864 \ Hz$
Comparing these with the given options,$844 \ Hz$ is not an integer multiple of $288 \ Hz$. Therefore,harmonics will not be produced at $844 \ Hz$.
146
MediumMCQ
The fundamental frequency of a sonometer wire is $n$. If the tension is increased $3$ times,the length is increased $3$ times,and the diameter is increased $2$ times,what will be the new frequency?
A
$2 n$
B
$\frac{\sqrt{3}}{2} n$
C
$\frac{n}{2 \sqrt{3}}$
D
$\sqrt{3} n$

Solution

(C) The fundamental frequency of a sonometer wire is given by the formula: $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length,$T$ is the tension,and $\mu$ is the linear mass density.
Since $\mu = \text{Area} \times \text{Density} = (\pi r^2) \rho = \pi (\frac{d}{2})^2 \rho = \frac{\pi d^2 \rho}{4}$,we can write $n \propto \frac{1}{L d} \sqrt{\frac{T}{\rho}}$.
Given the initial conditions: $n_1 = n$,$L_1 = L$,$T_1 = T$,$d_1 = d$.
New conditions: $L_2 = 3L$,$T_2 = 3T$,$d_2 = 2d$.
The new frequency $n_2$ is given by: $n_2 = n_1 \times (\frac{L_1}{L_2}) \times (\frac{d_1}{d_2}) \times \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $n_2 = n \times (\frac{L}{3L}) \times (\frac{d}{2d}) \times \sqrt{\frac{3T}{T}}$.
$n_2 = n \times \frac{1}{3} \times \frac{1}{2} \times \sqrt{3} = \frac{\sqrt{3}}{6} n = \frac{\sqrt{3}}{2 \times 3} n = \frac{n}{2 \sqrt{3}}$.
147
MediumMCQ
$A$ string $A$ has twice the length,twice the diameter,twice the tension,and twice the density of another string $B$. The overtone of $A$ which will have the same fundamental frequency as that of $B$ is:
A
first
B
second
C
third
D
fourth

Solution

(C) The fundamental frequency of a string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = \rho \pi r^2$.
For string $B$: $n_B = \frac{1}{2l_B} \sqrt{\frac{T_B}{\mu_B}}$.
Given for string $A$: $l_A = 2l_B$,$d_A = 2d_B \Rightarrow r_A = 2r_B$,$T_A = 2T_B$,and $\rho_A = 2\rho_B$.
Calculating $\mu_A$: $\mu_A = \rho_A \pi r_A^2 = (2\rho_B) \pi (2r_B)^2 = 8 \rho_B \pi r_B^2 = 8\mu_B$.
Now,the fundamental frequency of string $A$ is: $n_A = \frac{1}{2l_A} \sqrt{\frac{T_A}{\mu_A}} = \frac{1}{2(2l_B)} \sqrt{\frac{2T_B}{8\mu_B}} = \frac{1}{4} \left( \frac{1}{2l_B} \sqrt{\frac{T_B}{\mu_B}} \right) = \frac{1}{4} n_B$.
The frequency of the $p^{\text{th}}$ overtone of string $A$ is $f_p = (p+1)n_A$.
We want $f_p = n_B$,so $(p+1) \frac{n_B}{4} = n_B$.
This gives $p+1 = 4$,or $p = 3$.
Thus,the $3^{\text{rd}}$ overtone of string $A$ has the same frequency as the fundamental frequency of string $B$.
148
DifficultMCQ
$A$ sonometer wire is stretched by hanging a metal bob,the fundamental frequency of the wire is $n_1$. When the bob is completely immersed in water,the frequency of vibration of the wire becomes $n_2$. The relative density of the metal of the bob is
A
$\frac{n_1^2}{n_1^2-n_2^2}$
B
$\frac{n_2^2}{n_1^2-n_2^2}$
C
$\frac{n_1^2}{n_1^2+n_2^2}$
D
$\frac{n_2^2}{n_1^2+n_2^2}$

Solution

(A) The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension in the wire.
When the bob of weight $W$ is in air,the tension $T_1 = W$. Thus,$n_1 = \frac{1}{2L} \sqrt{\frac{W}{\mu}}$.
When the bob is immersed in water,the buoyant force $F_B$ acts on it,reducing the tension to $T_2 = W - F_B$. Thus,$n_2 = \frac{1}{2L} \sqrt{\frac{W - F_B}{\mu}}$.
Taking the ratio: $\frac{n_1}{n_2} = \sqrt{\frac{W}{W - F_B}} \implies \frac{n_1^2}{n_2^2} = \frac{W}{W - F_B}$.
Relative density $\sigma = \frac{W}{F_B}$.
Rearranging the ratio: $\frac{n_1^2}{n_2^2} = \frac{W}{W - (W/\sigma)} = \frac{\sigma}{\sigma - 1}$.
Solving for $\sigma$: $n_1^2(\sigma - 1) = n_2^2 \sigma \implies \sigma(n_1^2 - n_2^2) = n_1^2$.
Therefore,$\sigma = \frac{n_1^2}{n_1^2 - n_2^2}$.
149
MediumMCQ
When the tension in a string is increased by $3 \ kg \ wt$,the frequency of the fundamental mode increases in the ratio $2:3$. The initial tension in the string is: (in $kg \ wt$)
A
$1.6$
B
$2.0$
C
$2.4$
D
$2.8$

Solution

(C) The frequency of the fundamental mode of a stretched string is given by $f = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $T$ is the tension and $m$ is the linear mass density.
Let the initial tension be $T$. The initial frequency is $f = \frac{1}{2l} \sqrt{\frac{T}{m}}$.
When the tension is increased by $3 \ kg \ wt$,the new tension becomes $T' = T + 3$. The new frequency is $f' = \frac{1}{2l} \sqrt{\frac{T+3}{m}}$.
Given the ratio of frequencies is $f : f' = 2 : 3$,we have:
$\frac{f}{f'} = \sqrt{\frac{T}{T+3}} = \frac{2}{3}$
Squaring both sides:
$\frac{T}{T+3} = \frac{4}{9}$
$9T = 4(T + 3)$
$9T = 4T + 12$
$5T = 12$
$T = 2.4 \ kg \ wt$
Thus,the initial tension in the string is $2.4 \ kg \ wt$.
150
DifficultMCQ
$A$ sonometer wire is in unison with a tuning fork of frequency '$n$' when it is stretched by a weight of specific gravity '$d$'. When the weight is completely immersed in water,'$x$' beats are produced per second,then
A
$\frac{n}{n-x}=\frac{d}{d-1}$
B
$\frac{n}{n-x}=\sqrt{\frac{d}{d-1}}$
C
$\frac{n-x}{n}=\frac{d-1}{d}$
D
$\frac{n-x}{n}=\sqrt{\frac{d}{d-1}}$

Solution

(B) For a sonometer,the frequency of vibration $n$ is proportional to the square root of the tension $T$,i.e.,$n \propto \sqrt{T}$.
When the weight is in air,the tension $T_1 = V \cdot d \cdot \rho_w \cdot g$,where $V$ is the volume of the weight,$d$ is its specific gravity,and $\rho_w$ is the density of water.
Thus,$n \propto \sqrt{V \cdot d \cdot \rho_w \cdot g}$.
When the weight is immersed in water,the effective weight (tension) becomes $T_2 = V \cdot (d - 1) \cdot \rho_w \cdot g$ due to the buoyant force.
The new frequency $n'$ is given by $n' = n - x$ (since beats are produced).
Thus,$n' \propto \sqrt{V \cdot (d - 1) \cdot \rho_w \cdot g}$.
Taking the ratio of the two frequencies:
$\frac{n}{n-x} = \sqrt{\frac{V \cdot d \cdot \rho_w \cdot g}{V \cdot (d - 1) \cdot \rho_w \cdot g}}$
$\frac{n}{n-x} = \sqrt{\frac{d}{d-1}}$.

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