Waves and Sound JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Waves and Sound, free to read — no sign-in needed. The full chapter has 43 questions; sign in to attempt the remaining 38 in the exam simulator.

1. Q1JEE Advanced Adv 2023 (Paper 2)
   A string of length $1m$ and mass $2\times {10}^{-5}kg$ is under tension $T$. When the string vibrates, two successive harmonics are found to occur at frequencies $750Hz$ and $1000Hz$. The value of tension $T$ is _____ newton.
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   Answer: 5

   Given: $l=1m,m=2\times {10}^{–5}kg$. As successive harmonics are being given, so it is the case of both ends fixed. Now, ${f}_{n+1}-{f}_{n}=1000-750$ $\Rightarrow \dfrac{\left(n+1\right)}{2l}\sqrt{\dfrac{T}{\mu }}-\dfrac{n}{2l}\sqrt{\dfrac{T}{\mu }}=250$ $\Rightarrow \dfrac{1}{2l}\sqrt{\dfrac{T}{\mu }}=250$ $\Rightarrow \sqrt{\dfrac{T}{2\times {10}^{-5}}}=250\times 2\times 1$ $\Rightarrow \dfrac{T}{2\times {10}^{-5}}=25\times {10}^{4}$ $\Rightarrow T=50\times {10}^{-1}$ $\Rightarrow T=5N$
2. Q2JEE Advanced Adv 2023 (Paper 2)
   An ideal gas is in thermodynamic equilibrium. The number of degrees of freedom of a molecule of the gas is $n$. The internal energy of one mole of the gas is ${U}_{n}$ and the speed of sound in the gas is ${v}_{n}$. At a fixed temperature and pressure, which of the following is the correct option?
   1. A.${v}_{3}\lt {v}_{6}$ and ${U}_{3}\gt {U}_{6}$
   2. B.${v}_{5}\gt {v}_{3}$ and ${U}_{3}\gt {U}_{5}$
   3. C.${v}_{5}\gt {v}_{7}$ and ${U}_{5}\lt {U}_{7}$
   4. D.${v}_{6}\lt {v}_{7}$ and ${U}_{6}\lt {U}_{7}$
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   Answer: (C)

   As we know, internal energy of one mole of the gas ${U}_{n}=\dfrac{nRT}{2}$. Clearly, ${U}_{n}\propto n$, where $n$ is degree of freedom. Therefore, as $n$ increases, ${U}_{n}$ also increases. Now, speed of sound in gas ${v}_{n}=\sqrt{\dfrac{\gamma RT}{M}}=\sqrt{\dfrac{\left(1+\dfrac{2}{n}\right)RT}{M}}$ Clearly, as $n$ increases and ${v}_{n}$ decreases. Hence, ${U}_{7}\gt {U}_{5}$ and ${U}_{7}\gt {U}_{6}$ and ${v}_{5}\gt {v}_{7}$.
3. Q3JEE Advanced Adv 2020 (Paper 1)
   A stationary tuning fork is in resonance with an air column in a pipe. If the tuning fork is moved with a speed of $2m{s}^{-1}$ in front of the open end of the pipe and parallel to it, the length of the pipe should be changed for the resonance to occur with the moving tuning fork. If the speed of sound in air is $320m{s}^{-1},$ the smallest value of the percentage change required in the length of the pipe is ________.
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   Answer: 0.62

   $f=\dfrac{v}{4{L}_{1}}$ $f\left(\dfrac{320+0}{320-2}\right)=\dfrac{v}{4{L}_{2}}$ $\Rightarrow \dfrac{{L}_{2}}{{L}_{1}}=\dfrac{318}{320}$ $\Rightarrow \dfrac{{L}_{1}-{L}_{2}}{{L}_{1}}=1-\dfrac{318}{320}=\dfrac{2}{320}$ $\dfrac{\Delta L}{L}\times 100$
4. Q4JEE Advanced Adv 2018 (Paper 2)
   In an experiment to measure the speed of sound by a resonating air column, a tuning fork of frequency $500Hz$ is used. The length of the air column is varied by changing the level of water in the resonance tube. Two successive resonances are heard at air columns of length $50.7cmand83.9cm$ . Which of the following statements is (are) true?
   1. A.The speed of sound determined from this experiment is $332m{s}^{-1}$
   2. B.The end correction in this experiment is $0.9cm$
   3. C.The wavelength of the sound wave is $66.4cm$
   4. D.The resonance at $50.7cm$ corresponds to the fundamental harmonic
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   Answer: A,B,C

   Let ${n}_{1}$ harmonic is corresponding to $50.7cmand{n}_{2}$ harmonic is corresponding $83.9cm$ . since both one consecutive harmonics. $∴$ There difference $=\dfrac{\lambda }{2}$ $∴\dfrac{\lambda }{2}=\left(83.9-50.7\right)cm$ $\dfrac{\lambda }{2}=33.2cm$ $\lambda =66.4cm$ $∴\dfrac{\lambda }{4}=16.6cm$ length corresponding to fundamental mode must be close to $\dfrac{\lambda }{4}and50.7cm$ must be an odd multiple of this length $16.6\times 3=49.8cm$ . therefore $50.7is{3}^{rd}$ harmonic If end correction is e, then $e+50.7=\dfrac{3\lambda }{4}$ $e=49.8-50.7=0.9cm$ Speed of sound, $c=f\lambda$ $∴v=500\times 66.4cm/sec=332.000m/sec$
5. Q5JEE Advanced Adv 2017 (Paper 1)
   A stationary source emits sound of frequency ${f}_{0}=492Hz$ . The sound is reflected by a large car approaching the source with a speed of $2m{s}^{-1}$ . The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz? (Given that the speed of sound in air is 330 $m{s}^{-1}$ and the car reflects the sound at the frequency it has received).
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   Answer: 6

   Frequency of sound as received by large car approaching the source. ${f}_{1}=\dfrac{C+{V}_{0}}{C}{f}_{0}=\left(\dfrac{330+2}{330}\right)492Hz.$ This car now acts as source for reflected sound wave $∴{f}_{reflected}={f}_{1}$ Frequency of sound received by source, ${f}_{2}=\left(\dfrac{C}{C-{V}_{0}}\right){f}_{reflected}=\left(\dfrac{330}{330-2}\right)\times {f}_{1}=\dfrac{330}{328}\times \dfrac{332}{330}\times 492Hz$ $∴$ Beat frequency $=\left|{f}_{0}-{f}_{2}\right|=\left(\dfrac{332}{328}-1\right)\times 492Hz$ $=6Hz$