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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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An open pipe of length l vibrates in the fundamental mode. The pressure variation is maximum at :

1/4 from ends

The middle of the pipe

The ends of the pipe

At 1/8 from ends of pipe middle of the pipe

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Explanation

At the middle of pipe, node is formed.

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The fundamental frequency of pipe is 100 Hz and the other two frequencies are 300 Hz and 500 Hz then :

The pipe is open at both the ends

The pipe is closed at both the ends

One end open and another end is closed

None of the above

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Explanation

For closed organ pipe $n_{1} : n_{2} : n_{3} . . . = 1 : 3 : 5 : . . .$

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The fundamental frequency of an open pipe of length 0.5 m is equal to the frequency of the first overtone of a closed pipe of length l. The value of l_c is (m) :

1.5

0.75

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Explanation

First tone of open pipe = first overtone of closed pipe

⇒ $\frac{v}{2 l_{0}} = \frac{3 v}{4 l_{c}}$ ⇒ $l_{c} = \frac{3 \times 2 \times 0 .5}{4} = 0 .75 m$

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In a closed organ pipe, the frequency of the fundamental note is 50 Hz. The note of which of the following frequencies will not be emitted by it :

50 Hz

100 Hz

150 Hz

None of the above

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Explanation

Only odd harmonics are present.

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On producing the waves of frequency 1000 Hz in a Kundt's tube, the total distance between 6 successive nodes is 85 cm. Speed of sound in the gas filled in the tube is

330 m/s

340 m/s

350 m/s

300 m/s

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Explanation

Distance between six successive node

$= \frac{5 λ}{2} = 85 c m$ ⇒ $λ = \frac{2 \times 85}{5} = 34 c m = 0 .34 m$

Therefore speed of sound in gas

$= n λ = 1000 \times 0 .34 = 340 m / s$

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