Waves and Sound JEE Main previous year questions with solutions

4 solved JEE Main questions on Waves and Sound, free to read — no sign-in needed. The full chapter has 140 questions; sign in to attempt the remaining 136 in the exam simulator.

Answers checked against the official answer keys.

1. Q1JEE Main 2026 (21 Jan, Shift 1)Basics of Mechanical Waves
   Two strings $(A, B)$ having linear densities $\mu_{A}=2 \times 10^{-4} \mathrm{~kg} / \mathrm{m}$ and, $\mu_{B}=4 \times 10^{-4} \mathrm{~kg} / \mathrm{m}$ and lengths $L_{A}=2.5 \mathrm{~m}$ and $L_{B}=1.5 \mathrm{~m}$ respectively are joined. Free ends of $A$ and $B$ are tied to two rigid supports $C$ and $D$, respectively creating a tension of 500 N in the wire. Two identical pulses, sent from $C$ and $D$ ends, take time $t_{1}$ and $t_{2}$, respectively, to reach the joint. The ratio $t_{1} / t_{2}$ is :
   1. A.1.90
   2. B.1.67
   3. C.1.08
   4. D.1.18
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   Answer: (D)

   Wave velocity in string: $v = \sqrt{\frac{T}{\mu}}$ Time to reach joint: $t = \frac{L}{v} = L\sqrt{\frac{\mu}{T}}$ $t_1 = L_A\sqrt{\frac{\mu_A}{T}}$ and $t_2 = L_B\sqrt{\frac{\mu_B}{T}}$ $\frac{t_1}{t_2} = \frac{L_A}{L_B} \times \sqrt{\frac{\mu_A}{\mu_B}} = \frac{2.5}{1.5} \times \sqrt{\frac{2 \times 10^{-4}}{4 \times 10^{-4}}}$ $= \frac{5}{3} \times \sqrt{\frac{1}{2}} = \frac{5}{3\sqrt{2}} = \frac{5\sqrt{2}}{6} \approx 1.18$
2. Q2JEE Main 2025 (07 Apr, Shift 1)Beats
   Two harmonic waves moving in the same direction superimpose to form a wave $\mathrm{x}=\mathrm{a} \cos (1.5 \mathrm{t}) \cos (50.5 \mathrm{t})$ where $t$ is in seconds. Find the period with which they beat (close to nearest integer)
   1. A.6 s
   2. B.4 s
   3. C.1 s
   4. D.2 s
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   Answer: (D)

   The given equation can be written as $\mathrm{x}=\frac{\mathrm{a}}{2} \cos [1.5+50.5] \mathrm{t}+\frac{\mathrm{a}}{2} \cos [50.5-1.5]$ $x=\frac{a}{2} \cos [52 t]+\frac{a}{2} \cos [49 t]$ Here, $2 \pi \mathrm{f}_1 \& 2 \pi \mathrm{f}_2=49$ $\mathrm{f}_1=\frac{52}{2 \pi}, \mathrm{f}_2=\frac{49}{2 \pi}$ $\therefore \mathrm{f}_{\text {Bat }}=\mathrm{f}_1-\mathrm{f}_2=\frac{3}{2 \pi} \mathrm{~Hz}$ $\therefore \mathrm{T}_{\text {Beat }}=\frac{1}{\mathrm{f}_{\text {Beat }}}=\frac{2 \pi}{3} \mathrm{sec}$ $=2.09 \mathrm{sec} \approx 2 \mathrm{sec}$
3. Q3JEE Main 2024 (08 Apr, Shift 1)Organ Pipe (Vibration of Air Column)
   A closed and an open organ pipe have same lengths. If the ratio of frequencies of their seventh overtones is $\left(\frac{a-1}{a}\right)$ then the value of $\mathrm{a}$ is ________
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   Answer: 16

   For closed organ pipe $\mathrm{f}_{\mathrm{c}}=(2 \mathrm{n}+1) \frac{\mathrm{v}}{4 \ell}=\frac{15 \mathrm{v}}{4 \ell}$ For open organ pipe $\begin{aligned} & \mathrm{f}_{\mathrm{o}}=(\mathrm{n}+1) \frac{\mathrm{v}}{2 \ell}=\frac{8 \mathrm{v}}{2 \ell} \\ & \frac{\mathrm{f}_{\mathrm{c}}}{\mathrm{f}_{\mathrm{o}}}=\frac{15}{16}=\frac{\mathrm{a}-1}{\mathrm{a}} \\ & \Rightarrow \mathrm{a}=16\end{aligned}$
4. Q4JEE Main 2023 (31 Jan, Shift 2)Interference and Superposition of Waves
   The displacement equations of two interfering waves are given by ${y}_{1}=10\sin \left(\omega t+\dfrac{\pi }{3}\right)cm$, ${y}_{2}=5\left[\sin \left(\omega t\right)+\sqrt{3}\cos \omega t\right]cm$ respectively. The amplitude of the resultant wave is _____ $cm$.
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   Answer: 20

   Displacement equation of second wave can be written as,${y}_{2}=5\left(\sin \omega t+\sqrt{3}\cos \omega t\right) =10\left(\sin \omega t\times \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\times \cos \omega t\right) =10\left(\sin \omega t\times \cos \dfrac{\pi }{3}+\sin \dfrac{\pi }{3}\times \cos \omega t\right)$ $=10\sin \left(\omega t+\dfrac{\pi }{3}\right)$ As we can see, the phase difference between the waves is $0$. Therefore, resultant amplitude will be sum of the amplitudes of the waves. So $A={A}_{1}+{A}_{2}=10cm+10cm=20cm$