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📘 Waves and Sound — JEE Mains PYQs
Step-by-Step Solutions with Detailed Explanations
Master Waves and Sound for JEE Main with previous year questions on wave motion, beats, resonance, and organ pipes. Each solution is explained in a simple, step-by-step manner to boost your score.
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Waves and Sound — JEE Mains PYQs
JEE Main 2018–2025Q1. JEE Main 2025 (8 April Shift 2)
Two strings with circular cross section and made of same material, are stretched to have same amount of tension. A transverse wave is then made to pass through both the strings. The velocity of the wave in the first string having the radius of cross section R is \(v_1\), and that in the other string having radius of cross section \(R/2\) is \(v_2\). Then \(v_2/v_1 =\) :
(1) \(\sqrt{2}\) (2) 2 (3) 8 (4) 4
(1) \(\sqrt{2}\) (2) 2 (3) 8 (4) 4
✅ Answer & Explanation:
Correct Option: (2)
- The velocity of a transverse wave in a string is \(v = \sqrt{T/\mu}\), where \(\mu\) is the mass per unit length. Since \(\mu = \rho \cdot \text{Area} = \rho \pi R^2\), we have \(v \propto \frac{1}{R}\).
- Therefore, \(\frac{v_2}{v_1} = \frac{R_1}{R_2} = \frac{R}{\frac{R}{2}} = 2\).
Q2. JEE Main 2025 (4 April Shift 2)
Displacement of a wave is expressed as \(x(t) = 5 \cos(628t + \frac{\pi}{2}) \, \text{m}\). The wavelength of the wave when its velocity is 300 m/s is :
(1) 5 m (2) 3 m (3) 0.5 m (4) 0.33 m
(1) 5 m (2) 3 m (3) 0.5 m (4) 0.33 m
✅ Answer & Explanation:
Correct Option: (2)
- From the equation, angular frequency \(\omega = 628 \, \text{rad/s}\).
- Frequency \(f = \frac{\omega}{2\pi} = \frac{628}{2 \times 3.14} = 100 \, \text{Hz}\).
- Using \(v = f\lambda\), we get \(\lambda = \frac{v}{f} = \frac{300}{100} = 3 \, \text{m}\).
Q3. JEE Main 2025 (28 Jan Shift 1)
Assertion A: A sound wave has higher speed in solids than gases.
Reason R: Gases have higher value of Bulk modulus than solids.
(1) Both A and R are true but R is NOT the correct explanation of A
(2) A is true but R is false
(3) A is false but R is true
(4) Both A and R are true and R is the correct explanation of A
Reason R: Gases have higher value of Bulk modulus than solids.
(1) Both A and R are true but R is NOT the correct explanation of A
(2) A is true but R is false
(3) A is false but R is true
(4) Both A and R are true and R is the correct explanation of A
✅ Answer & Explanation:
Correct Option: (2)
- Assertion: True. The speed of sound is higher in solids because they have much larger elastic moduli compared to gases.
- Reason: False. Solids have significantly higher Bulk modulus than gases.
Q4. JEE Main 2022 (26 Jul Shift 2)
A transverse wave is represented by \(y = 2 \sin(\omega t - kx) \, \text{cm}\). The value of wavelength (in cm) for which the wave velocity becomes equal to the maximum particle velocity, will be :
(1) \(4\pi\) (2) \(2\pi\) (3) \(\pi\) (4) 2
(1) \(4\pi\) (2) \(2\pi\) (3) \(\pi\) (4) 2
✅ Answer & Explanation:
Correct Option: (1)
- Wave velocity \(v_w = \omega / k\).
- Maximum particle velocity \(v_{p,max} = A\omega\).
- Given \(v_w = v_{p,max} \implies \frac{\omega}{k} = 2\omega \implies k = \frac{1}{2}\).
- Since \(k = 2\pi / \lambda\), then \(\frac{2\pi}{\lambda} = \frac{1}{2} \implies \lambda = 4\pi \, \text{cm}\).
Q5. JEE Main 2021 (31 Aug Shift 1)
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 N. The wire resonates at a frequency of 500 Hz. The next higher frequency at which the same wire resonates is 550 Hz. The length of the wire is ____ m.
✅ Answer & Explanation:
Correct Answer: 10
- Fundamental frequency \(f_0 = \Delta f = 550 - 500 = 50 \, \text{Hz}\).
- Velocity \(v = \sqrt{T/\mu} = \sqrt{\frac{900}{9 \times 10^{-4}}} = \sqrt{10^6} = 1000 \, \text{m/s}\).
- For fixed ends, \(f_0 = \frac{v}{2L} \implies 50 = \frac{1000}{2L} \implies 100L = 1000 \implies L = 10 \, \text{m}\).
Q6. JEE Main 2021 (20 Jul Shift 1)
The amplitude of wave disturbance propagating in the positive x-direction is given by \(y = \frac{1}{(1+x)^2}\) at time \(t = 0\) and \(y = \frac{1}{1+(x-2)^2}\) at \(t = 1 \, \text{s}\). The velocity of the wave will be _____ m/s.
✅ Answer & Explanation:
Correct Answer: 2
- A wave propagating in the +x direction follows the form \(f(x - vt)\).
- Comparing the positions of the peaks: At \(t=0\), peak is at \(x=0\). At \(t=1\), peak is at \(x-vt = x-2 \implies vt = 2\).
- Since \(t=1\), velocity \(v = 2 \, \text{m/s}\).
Q7. JEE Main 2025 (7 April Shift 1)
Two harmonic waves moving in the same direction superimpose to form a wave \(x = a \cos(1.5t) \cos(50.5t)\) where \(t\) is in seconds. Find the period with which they beat (close to nearest integer) :
(1) 6 s (2) 4 s (3) 1 s (4) 2 s
(1) 6 s (2) 4 s (3) 1 s (4) 2 s
✅ Answer & Explanation:
Correct Option: (4)
- Using the identity \(2 \cos A \cos B = \cos(A-B) + \cos(A+B)\), the resultant wave is the sum of two waves with angular frequencies \(\omega_1 = 50.5 - 1.5 = 49 \, \text{rad/s}\) and \(\omega_2 = 50.5 + 1.5 = 52 \, \text{rad/s}\).
- Beat angular frequency: \(\omega_{beat} = \omega_2 - \omega_1 = 3 \, \text{rad/s}\).
- Beat period: \(T = \frac{2\pi}{\omega_{beat}} = \frac{2 \times 3.14}{3} \approx 2.09 \, \text{s}\). The nearest integer is 2 s.
Q8. JEE Main 2025 (3 April Shift 2)
In the resonance experiment, two air columns (closed at one end) of 100 cm and 120 cm long, give 15 beats per second when each one is sounding in the respective fundamental modes. The velocity of sound in the air column is :
(1) 335 m/s (2) 370 m/s (3) 340 m/s (4) 360 m/s
(1) 335 m/s (2) 370 m/s (3) 340 m/s (4) 360 m/s
✅ Answer & Explanation:
Correct Option: (4)
- For a closed organ pipe, fundamental frequency \(f = \frac{v}{4L}\).
- Beat frequency: \(f_1 - f_2 = \frac{v}{4L_1} - \frac{v}{4L_2} = 15\).
- \(\frac{v}{4} \left( \frac{1}{1.0} - \frac{1}{1.2} \right) = 15 \implies \frac{v}{4} \left( \frac{0.2}{1.2} \right) = 15\).
- \(\frac{v}{4} \times \frac{1}{6} = 15 \implies v = 15 \times 24 = 360 \, \text{m/s}\).
Q9. JEE Main 2022 (28 Jun Shift 1)
The velocity of sound in a gas, in which two wavelengths 4.08 m and 4.16 m produce 40 beats in 12 s, will be :
(1) 282.8 m/s (2) 175.5 m/s (3) 353.6 m/s (4) 707.2 m/s
(1) 282.8 m/s (2) 175.5 m/s (3) 353.6 m/s (4) 707.2 m/s
✅ Answer & Explanation:
Correct Option: (4)
- Beat frequency \(f_b = \frac{40}{12} = \frac{10}{3} \, \text{Hz}\).
- \(f_1 - f_2 = \frac{v}{\lambda_1} - \frac{v}{\lambda_2} = \frac{10}{3}\).
- \(v \left( \frac{1}{4.08} - \frac{1}{4.16} \right) = \frac{10}{3} \implies v \left( \frac{0.08}{4.08 \times 4.16} \right) = \frac{10}{3}\).
- Solving for \(v\) gives 707.2 m/s.
Q10. JEE Main 2022 (26 Jun Shift 2)
A set of 20 tuning forks is arranged in a series of increasing frequencies. If each fork gives 4 beats with respect to the preceding fork and the frequency of the last fork is twice the frequency of the first, then the frequency of last fork is _____ Hz.
✅ Answer & Explanation:
Correct Answer: 152
- This forms an Arithmetic Progression: \(f_n = f_1 + (n-1)d\).
- Given \(n=20\), \(d=4\), and \(f_{20} = 2f_1\).
- \(2f_1 = f_1 + (19 \times 4) \implies f_1 = 76 \, \text{Hz}\).
- Frequency of the last fork \(f_{20} = 2 \times 76 = 152 \, \text{Hz}\).
Q11. JEE Main 2020 (05 Sep Shift 1)
In a resonance tube experiment when the tube is filled with water up to a height of 17.0 cm from bottom, it resonates with a given tuning fork. When the water level is raised the next resonance with the same tuning fork occurs at a height of 24.5 cm. If the velocity of sound in air is 330 m/s, the tuning fork frequency is :
(1) 2200 Hz (2) 550 Hz (3) 1100 Hz (4) 3300 Hz
(1) 2200 Hz (2) 550 Hz (3) 1100 Hz (4) 3300 Hz
✅ Answer & Explanation:
Correct Option: (1)
- Wait, the heights given are from the bottom. The air column lengths are \(L_1 = L_{total} - 17.0\) and \(L_2 = L_{total} - 24.5\).
- The difference in air column lengths is \(\Delta L = L_1 - L_2 = 24.5 - 17.0 = 7.5 \, \text{cm} = 0.075 \, \text{m}\).
- In a resonance tube, the distance between consecutive resonances is \(\lambda/2 = \Delta L\).
- \(\lambda = 2 \times 0.075 = 0.15 \, \text{m}\).
- \(f = \frac{v}{\lambda} = \frac{330}{0.15} = 2200 \, \text{Hz}\).
Q12. JEE Main 2025 (22 Jan Shift 1)
A closed organ and an open organ tube are filled by two different gases having same bulk modulus but different densities \(\rho_1\) and \(\rho_2\), respectively. The frequency of 9th harmonic of closed tube is identical with 4th harmonic of open tube. If the length of the closed tube is 10 cm and the density ratio of the gases is \(\rho_1 : \rho_2 = 1 : 16\), then the length of the open tube is :
(1) \(\frac{15}{7} \, \text{cm}\) (2) \(\frac{20}{7} \, \text{cm}\) (3) \(\frac{15}{9} \, \text{cm}\) (4) \(\frac{20}{9} \, \text{cm}\)
(1) \(\frac{15}{7} \, \text{cm}\) (2) \(\frac{20}{7} \, \text{cm}\) (3) \(\frac{15}{9} \, \text{cm}\) (4) \(\frac{20}{9} \, \text{cm}\)
✅ Answer & Explanation:
Correct Option: (4)
- Velocity in gas: \(v = \sqrt{B/\rho}\). Ratio \(\frac{v_1}{v_2} = \sqrt{\frac{\rho_2}{\rho_1}} = \sqrt{\frac{16}{1}} = 4\).
- 9th harmonic of closed pipe: \(f_C = 9 \left( \frac{v_1}{4L_C} \right)\).
- 4th harmonic of open pipe: \(f_O = 4 \left( \frac{v_2}{2L_O} \right) = \frac{2v_2}{L_O}\).
- Equating: \(\frac{9v_1}{40} = \frac{2v_2}{L_O} \implies L_O = \frac{80v_2}{9v_1} = \frac{80}{9(4)} = \frac{20}{9} \, \text{cm}\).
Q13. JEE Main 2022 (25 Jun Shift 1)
The first overtone frequency of an open organ pipe is equal to the fundamental frequency of a closed organ pipe. If the length of the closed organ pipe is 20 cm. The length of the open organ pipe is _____ cm.
✅ Answer & Explanation:
Correct Answer: 80
- Fundamental frequency of closed pipe \(f_C = \frac{v}{4L_C}\).
- First overtone of open pipe \(f_{O,1} = 2 \left( \frac{v}{2L_O} \right) = \frac{v}{L_O}\).
- Equating: \(\frac{v}{L_O} = \frac{v}{4L_C} \implies L_O = 4L_C = 4 \times 20 = 80 \, \text{cm}\).
Q14. JEE Main 2021 (26 Aug Shift 1)
Two travelling waves produces a standing wave represented by equation \(y = (1.0 \, \text{mm}) \cos[(1.57 \, \text{cm}^{-1})x] \sin[(78.5 \, \text{s}^{-1})t]\). The node closest to the origin in the region \(x > 0\) will be at \(x =\) _____ cm.
✅ Answer & Explanation:
Correct Answer: 1
- Nodes occur where the amplitude part \(\cos(kx) = 0\).
- \(1.57x = \pi/2 \implies (\pi/2)x = \pi/2 \implies x = 1 \, \text{cm}\).
- (Note: \(1.57 \approx \frac{3.14}{2}\)).
📘 Exam Preparation Tip for Waves and Sound
- Wave Velocity: For a string, \(v = \sqrt{T/\mu}\), where \(\mu\) is mass per unit length. For sound, \(v = \sqrt{B/\rho}\) in fluids and \(v = \sqrt{Y/\rho}\) in solids.
- Wave Equation: \(y = A \sin(\omega t - kx)\) represents a wave traveling in the +x direction. The wave number \(k = 2\pi/\lambda\) and angular frequency \(\omega = 2\pi f\).
- Beats: When two waves of slightly different frequencies superpose, the beat frequency is \(f_b = |f_1 - f_2|\). Beat period is \(T_b = 1/f_b\).
- Standing Waves: In a string fixed at both ends, allowed frequencies are \(f = \frac{nv}{2L}\). Nodes are points of zero displacement, and antinodes are points of maximum displacement.
- Organ Pipes: For open pipe: \(f = \frac{nv}{2L}\) (n = 1,2,3...). For closed pipe: \(f = \frac{(2n-1)v}{4L}\) (n = 1,2,3...). Only odd harmonics are present in a closed pipe.
- Resonance Tube: The distance between consecutive resonances is \(\lambda/2\). Use this to find the wavelength and frequency.
- Particle vs Wave Velocity: In a wave, the particle velocity is \(v_p = \partial y/\partial t = -A\omega \cos(\omega t - kx)\) for a sine wave. The maximum particle velocity is \(v_{p,max} = A\omega\).