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Waves MCQs for NEET — Physics Questions with Answers

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A cylindrical tube, open at both ends, has a fundamental frequency f₀ in air. The tube is dipped vertically into water such that half of its length is inside water. The fundamental frequency of the air column now is

3 f₀/4

f₀

f₀/2

2 f₀

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Explanation

$n_{open} = \frac{v}{2 l_{open}}$

$n_{closed} = \frac{v}{4 l_{closed}} = \frac{v}{4 l_{open} / 2} = \frac{v}{2 l_{open}}$

$(A s l_{c l o s e d} = \frac{l_{o p e n}}{2})$ , i.e. frequency remains unchanged.

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If the length of a closed organ pipe is 1.5 m and the velocity of sound is 330 m/s, then the frequency for the second note is

220 Hz

165 Hz

110 Hz

55 Hz

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Explanation

For closed pipe second note = $\frac{3 v}{4 l} = \frac{3 \times 330}{4 \times 1 .5} = 165 H z$ .

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A pipe 30 cm long is open at both ends. Which harmonic mode of the pipe is resonantly excited by a 1.1 kHz source? (Take the speed of sound in air = 330 ms^–1)

First

Second

Third

Fourth

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Explanation

Fundamental frequency of open pipe

$n_{1} = \frac{v}{2 l} = \frac{330}{2 \times 0 .3} = 550 H z$

Second harmonic = $2 \times n_{1} = 1100 H z = 1 .1 k H z$

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A source of sound placed at the open end of a resonance column sends an acoustic wave of pressure amplitude $ρ_{0}$ inside the tube. If the atmospheric pressure is $ρ_{A}$ , then the ratio of maximum and minimum pressure at the closed end of the tube will be :

$\frac{(ρ_{A} + ρ_{0})}{(ρ_{A} - ρ_{0})}$

$\frac{(ρ_{A} + 2 ρ_{0})}{(ρ_{A} - 2 ρ_{0})}$

$\frac{ρ_{A}}{ρ_{A}}$

$\frac{(ρ_{A} + \frac{1}{2} ρ_{0})}{(ρ_{A} - \frac{1}{2} ρ_{0})}$

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Explanation

Maximum pressure at closed-end will be atmospheric pressure adding with acoustic wave pressure

So $ρ_{\max} = ρ_{A} + ρ_{0}$ and $ρ_{\min} = ρ_{A} - ρ_{0}$

Thus $\frac{ρ_{\max}}{ρ_{\min}} = \frac{ρ_{A} + ρ_{0}}{ρ_{A} - ρ_{0}}$

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Two closed pipe produce 10 beats per second when emitting their fundamental nodes. If their length are in ratio of 25 : 26. Then their fundamental frequency in Hz, are :

270, 280

260, 270

260, 250

260, 280

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Explanation

$n_{1} - n_{2} = 10$ ..…(i)

Using $n_{1} = \frac{v}{4 l_{1}}$ and $n_{2} = \frac{v}{4 l_{2}}$

⇒ $\frac{n_{1}}{n_{2}} = \frac{l_{2}}{l_{1}} = \frac{26}{25}$ …..(ii)

After solving these equation $n_{1} = 260 H z$ , $n_{2} = 250 H z$

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If v is the speed of sound in the air then the shortest length of the closed pipe which resonates to a frequency n :

$\frac{v}{4 n}$

$\frac{v}{2 n}$

$\frac{2 n}{v}$

$\frac{4 n}{v}$

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Explanation

For shortest length of pipe mode of vibration must be fundamental i.e., $n = \frac{v}{4 l}$ ⇒ $l = \frac{v}{4 n} .$

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The frequency of fundamental tone in an open organ pipe of length 0.48 m is 320 Hz. The speed of sound is 320 m/sec. Frequency of fundamental tone in closed organ pipe will be :

153.8 Hz

160.0 Hz

320.0 Hz

143.2 Hz

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Explanation

$n_{Closed} = \frac{1}{2} (n_{Open}) = \frac{1}{2} \times 320 = 160 H z$

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What is the minimum length of a tube, open at both ends, that resonates with tuning fork of frequency 350 Hz? [velocity of sound in air = 350 m/s]

50 cm

100 cm

75 cm

25 cm

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Explanation

Fundamental frequency $n = \frac{v}{2 l}$

$\Rightarrow 350 = \frac{350}{2 L}$ $\Rightarrow L = \frac{1}{2} m = 50 c m .$

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The harmonics which are present in a pipe open at one end are :

Odd harmonics

Even harmonics

Even as well as odd harmonics

None of these

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Explanation

In closed pipe only odd harmonics are present

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The stationary wave $y = 2 a \sin k x \cos ω t$ in a closed organ pipe is the result of the superposition of $y = a \sin (ω t - k x)$ and

$y = - a \cos (ω t + k x)$

$y = - a \sin (ω t + k x)$

$y = a \sin (ω t + k x)$

$y = a \cos (ω t + k x)$

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Explanation

In closed organ pipe. If $y_{i n c i d e n t} = a \sin (ω t - k x)$ then $y_{r e f l e c t e d} = a \sin (ω t + k x + π) = - a \sin (ω t + k x)$

Superimposition of these two waves give the required stationary wave.

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