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Waves MCQs for NEET — Physics Questions with Answers

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When a tuning fork A of unknown frequency is sounded with another tuning fork B of frequency 256 Hz, then 3 beats per second are observed. After that A is loaded with wax and sounded, the again 3 beats per second are observed. The frequency of the tuning fork A is :

250 Hz

253 Hz

259 Hz

262 Hz

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Explanation

It is given that :

n_A = Unknown frequency = ?

n_B = Known frequency = 256 Hz

x = 3 bps, which remains same after loading

Unknown tuning fork A is loaded so n_A↓

Hence n_A↓ – n_B = x ... (i) → Correct

n_B – n_A↓ = x ... (ii) → Wrong

⇒ n_A = n_B + x = 256 + 3 = 259 Hz.

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When two sound waves are superimposed, beats are produced when they have :

Different amplitudes and phases

Different velocities

Different phases

Different frequencies

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Explanation

For producing beats, their must be small difference in frequency.

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A tuning fork A of frequency 200 Hz is sounded with fork B, the number of beats per second is 5. By putting some wax on A, the number of beats increases to 8. The frequency of fork B is :

200 Hz

195 Hz

192 Hz

205 Hz

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Explanation

n_A↓ – n_B = x↑ ... (i) → Wrong

n_B – n_A↓ = x↑ ... (ii) → Correct

⇒ n_B = n_A + x = 200 + 5 = 205 Hz.

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Two tuning forks have frequencies 380 and 384 Hz respectively. When they are sounded together, they produce 4 beats. After hearing the maximum sound, how long will it take to hear the minimum sound?

$\frac{1}{2}$ sec

$\frac{1}{4}$ sec

$\frac{1}{8}$ sec

$\frac{1}{16}$ sec

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Explanation

Beat period $T = \frac{1}{n_{1} ~ n_{2}} = \frac{1}{384 - 380} = \frac{1}{4} \sec$ .

Hence minimum time interval between maxima and minima $t = \frac{T}{2} = \frac{1}{8} \sec .$

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A couple of tuning forks produces 2 beats in the time interval of 0.4 seconds. So the beat frequency is :

8 Hz

5 Hz

2 Hz

10 Hz

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Explanation

Beat frequency = $\frac{2}{0 .4} = 5 H z$

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It is possible to hear beats from the two vibrating sources of frequency :

100 Hz and 150 Hz

20 Hz and 25 Hz

400 Hz and 500 Hz

1000 Hz and 1500 Hz

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Explanation

For hearing beats, difference of frequencies should be approximately 10 Hz.

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Two sound waves of wavelengths 5m and 6m formed 30 beats in 3 seconds. The velocity of sound is :

300 ms^–1

310 ms^–1

320 ms^–1

330 ms^–1

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Explanation

No of beats, x = $Δ n = \frac{30}{3} = 10 H z$

⇒ Also $Δ n = v [\frac{1}{λ_{1}} - \frac{1}{λ_{2}}] = v [\frac{1}{5} - \frac{1}{6}]$ =10⇒ v = 300 m/s

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Two sound sources when sounded simultaneously produce four beats in 0.25 seconds. The difference in their frequencies must be :

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Explanation

No. of beats = frequency difference = $\frac{4}{0 .25} = 16$

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Two strings X and Y of a sitar produce a beat frequency 4 Hz. When the tension of the string Y is slightly increased the beat frequency is found to be 2 Hz. If the frequency of X is 300 Hz, then the original frequency of Y was :

296 Hz

298 Hz

302 Hz

304 Hz

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Explanation

$n_{x} = 300 H z, n_{y} = ?$

x = beat frequency = 4 Hz, which is decreasing (4 → 2)

after increasing the tension of the string y.

Also tension of wire y increasing so $n_{y} ↑$ $(∵ n \propto \sqrt{T})$

Hence $n_{x} - n_{y} ↓ = x ↓$ → Correct

$n_{y} ↑ - n_{x} = x ↓$ → Wrong

⇒ $n_{y} = n_{x} - x = 300 - 4 = 296 H z$

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Two vibrating tuning forks produce progressive waves given by $Y_{1} = 4 \sin 500 π t$ and $Y_{2} = 2 \sin 506 π t .$ Number of beats produced per minute is :

360

180

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Explanation

From the given equations of progressive waves $ω_{1} = 500 π$ and $ω_{2} = 506 π$

∴ $n_{1} = 250$ and $n_{2} = 253$

So beat frequency $= n_{2} - n_{1} = 253 - 250 = 3$ beats per sec

∴ Number of beats per min = 180.

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