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 139 questions; sign in to attempt the remaining 135 in the exam simulator.

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 $A={A}_{1}+{A}_{2}=10cm+10cm=20cm$