Sound Waves – JEE Main Previous Year Questions with Solutions
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Q. Three sound waves of equal amplitudes have frequencies (–1), , (+1). They superpose to give beats. The number of beats produced per second will be :- (1) 2                   (2) 1               (3) 4                  (4) 3 [AIEEE – 2009]

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Sol. (1)

Q. A motor cycle starts from rest and accelerates along a straight path at 2 \mathrm{m} / \mathrm{s}^{2}. At the starting point of the motor cycle there is a stationary electric siren. How far has the motor cycle gone when the driver hears the frequency of the siren at 94% of its value when the motor cycle was at rest ? (Speed of sound = 330 \mathrm{ms}^{-1}) :- (1) 147 m               (2) 196 m               (3) 49 m               (4) 98 m [AIEEE – 2009]

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Sol. (4) $n^{\prime}=\frac{v-v_{0}}{v} n=\frac{94}{100} n$ from (iii) eq”. of motion $v^{2}=u^{2}+2 a s$ $\Rightarrow \mathrm{v}_{0}^{2}=0+2 \mathrm{as} \Rightarrow \mathrm{v}_{0}=\sqrt{2 \mathrm{as}}$ $\Rightarrow \frac{94}{100} \mathrm{n}=\left(\frac{\mathrm{v}-\sqrt{2 \mathrm{as}}}{\mathrm{v}}\right) \mathrm{n} \Rightarrow \mathrm{s}=98 \mathrm{m}$

Q. A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s. (1) 6             (2) 4               (3) 12                (4) 8 [JEE Mains – 2014]

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Sol. (1) Fundamental frequency of closed organ pipe $\mathrm{f}_{0}=\frac{\mathrm{v}}{4 \ell}=\frac{340}{4 \times 0.85}=100 \mathrm{Hz}$ So possible frequencies below 1250 Hz are 100 Hz, 300 Hz, 500 Hz, 700 Hz, 900 Hz, 1100 Hz $\Rightarrow$ No. of frequencies $=6$

Q. A train is moving on a straight track with speed 20 \mathrm{ms}^{-1}. It is blowing its whistle at the frequency of 1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound = 320 \mathrm{ms}^{-1}) close to :- (1) 18%           (2) 24%               (3) 6%               (4) 12% [JEE Mains – 2015]

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Sol. (4) $\mathrm{f}^{\prime}=\mathrm{f}\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right)$ $\mathrm{f}^{\prime \prime}=\mathrm{f}\left(\frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right)$ $\frac{\mathrm{f}^{\prime \prime}}{\mathrm{f}^{\prime}}=\frac{\mathrm{v}-\mathrm{v}_{\mathrm{S}}}{\mathrm{v}+\mathrm{v}_{\mathrm{S}}}$ $\frac{\mathrm{f}^{\prime \prime}-\mathrm{f}^{\prime}}{\mathrm{f}^{\prime}} \times 100=\frac{\mathrm{v}-\mathrm{v}_{\mathrm{S}}-\mathrm{v}-\mathrm{v}_{\mathrm{S}}}{\mathrm{v}+\mathrm{v}_{\mathrm{S}}} \times 10$ $=-\frac{2 \mathrm{v}_{\mathrm{S}}}{\mathrm{v}+\mathrm{v}_{\mathrm{S}}}=-\frac{2}{1+\frac{\mathrm{v}}{\mathrm{v}_{\mathrm{S}}}} \times 100 \simeq 12 \%$

Q. An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer? (speed of light $=3 \times 10^{8} \mathrm{ms}^{-1}$ ) (1) 17.3 GHz (2) 15.3 GHz (3) 10.1 GHz (4) 12.1 GHz [JEE Mains – 2017]

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Sol. (1) Doppler effect in light (speed of observer is not very small compare to speed of light) $\mathrm{f}^{1}=\sqrt{\frac{1+\mathrm{V} / \mathrm{C}}{1-\mathrm{V} / \mathrm{C}}} \mathrm{f}_{\mathrm{source}}=\sqrt{\frac{1+1 / 2}{1-1 / 2}}(10 \mathrm{GHz})$ = 17.3 GHz

Q. A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is $2.7 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}$ and its Young’s modulus is $9.27 \times 10^{10}$ Pa. What will be the fundamental frequency of the longitudinal vibrations ? (1) 2.5 kHz (2) 10 kHz (3) 7.5 kHz (4) 5 kHz [JEE Mains – 2018]

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Sol. (4) Velocity of wave $=\sqrt{\frac{\mathrm{Y}}{\rho}}$ $=\sqrt{\frac{9.27 \times 10^{10}}{2.7 \times 10^{3}}}$ $=\sqrt{3.433 \times 10^{7}}$ $=10^{3} \times \sqrt{34.33}$ $\mathrm{v}_{\omega}=5.85 \times 10^{3} \mathrm{m} / \mathrm{sec}$ Since rod is clamped at middle fundamental wave shape is as follow $\frac{\lambda}{2}=\mathrm{L}$ $\lambda=2 \mathrm{L}$ $\mathrm{L}=60 \mathrm{cm}=0.6 \mathrm{m}(\mathrm{given})$ $\lambda=1.2 \mathrm{m}$ $\mathrm{v}=\mathrm{f} \lambda$ $\mathrm{f}=\frac{\mathrm{v}}{\lambda}=\frac{5.85 \times 10^{3}}{1.2}$ $=4.88 \times 10^{3} \mathrm{Hz} \square 5 \mathrm{KHz}$

• February 14, 2021 at 12:13 pm

Ty

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• November 16, 2020 at 8:15 pm

Sir … Pls keep more n more questions for reference and keep the important questions from all the previous years……
Thanking you sir …

3
• September 28, 2020 at 4:40 pm

Plz keep more questions

7
• August 27, 2020 at 2:45 pm

Thank you sir.

7
• July 19, 2020 at 6:24 am

These were easy questions.

4