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Waves MCQs for NEET — Physics Questions with Answers

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In one metre long open pipe what is the harmonic of resonance obtained with a tuning fork of frequency 480 Hz :

First

Second

Third

Fourth

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Explanation

In case of open pipe, $n = \frac{N}{2 l}$ where N = order of harmonics = order of mode of vibration

⇒ $N = \frac{n \times 2 l}{v}$

$= \frac{480}{330} \times 2 \times 1$ = 3(Here v = 330 m/s)

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In a resonance pipe the first and second resonances are obtained at depths 22.7 cm and 70.2 cm respectively. What will be the end correction

1.05 cm

115.5 cm

92.5 cm

113.5 cm

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Explanation

For end correction x, $\frac{l_{2} + x}{l_{1} + x} = \frac{3 λ / 4}{λ / 4} = 3$

$x = \frac{l_{2} - 3 l_{1}}{2} = \frac{70 .2 - 3 \times 22 .7}{2} = 1 .05 c m$

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An open tube is in resonance with string (frequency of vibration of the tube is n₀). If the tube is dipped in water so that 75% of the length of the tube is inside water, then the ratio of the frequency of tube to string now will be :

$\frac{2}{3}$

$\frac{3}{2}$

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Explanation

For open tube, $n_{0} = \frac{v}{2 l}$

For closed tube length available for resonance is

$l' = l \times \frac{25}{100} = \frac{l}{4}$

∴ Fundamental frequency of water filled tube

$n = \frac{v}{4 l'} = \frac{v}{4 \times (l / 4)} = \frac{v}{l} = 2 n_{0}$ ⇒ $\frac{n}{n_{0}} = 2$

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A source of sound of frequency 450 cycles/sec is moving towards a stationary observer with 34 m/sec speed. If the speed of sound is 340 m/sec, then the apparent frequency will be

410 cycles/sec

500 cycles/sec

550 cycles/sec

450 cycles/sec

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Explanation

$n' = n (\frac{v}{v - v_{0}}) = 450 (\frac{340}{340 - 34}) = 500 c y c l e s / \sec$

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The frequency of a whistle of an engine is 600 cycles/sec is moving with the speed of 30 m/sec towards an observer. The apparent frequency will be (velocity of sound = 330 m/s) [MP PMT 1989]

600 cps

660 cps

990 cps

330 cps

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Explanation

$n' = n (\frac{v}{v - v_{S}}) = 600 (\frac{330}{300}) = 660 c p s$

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An observer moves towards a stationary source of sound of frequency n. The apparent frequency heard by him is 2n. If the velocity of sound in air is 332 m/sec, then the velocity of the observer is :

166 m/sec

664 m/sec

332 m/sec

1328 m/sec

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Explanation

$n' = n (\frac{v + v_{O}}{v}) \Rightarrow 2 n = n (\frac{v + v_{0}}{v}) \Rightarrow \frac{v + v_{0}}{v} = 2$

$\Rightarrow v_{O} = v = 332$ m/sec

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A source of sound emitting a note of frequency 200 Hz moves towards an observer with a velocity v equal to the velocity of sound. If the observer also moves away from the source with the same velocity v, the apparent frequency heard by the observer is

50 Hz

100 Hz

150 Hz

200 Hz

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Explanation

Since there is no relative motion between observer and source, therefore there is no apparent change in frequency.

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A source of sound is traveling towards a stationary observer. The frequency of the sound heard by the observer is three times the original frequency. The velocity of sound is v m/sec. The speed of the source will be :

$\frac{2}{3} v$

$\frac{3}{2} v$

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Explanation

$n^{'} = n (\frac{v}{v - v_{S}})$ ⇒ $\frac{n'}{n} = \frac{v}{v - v_{S}}$

$\Rightarrow \frac{v}{v - v_{S}} = 3 \Rightarrow v_{s} = \frac{2 v}{3}$

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A sound source is moving towards a stationary observer with 1/10 of the speed of sound. The ratio of apparent to real frequency is :

10/9

11/10

(11/10)²

(9/10)²

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Explanation

$n' = n (\frac{v}{v - v_{S}}) = n (\frac{v}{v - v / 10})$

⇒ $\frac{n'}{n} = \frac{10}{9}$

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The speed of sound in air at a given temperature is 350 m/s. An engine blows whistle at a frequency of 1200 cps. It is approaching the observer with velocity 50 m/s. The apparent frequency in cps heard by the observer will be :

600

1050

1400

2400

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Explanation

$n'=n (\frac{v}{v - v_{s}}) = 1200 \times (\frac{350}{350 - 50})$ =1400 cps

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