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What is the base frequency if a pipe gives notes of frequencies 425, 255 and 595 and decide whether it is closed at one end or open at both ends :

17, closed

85, closed

17, open

85, open

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Explanation

Let the base frequency be n for closed pipe then notes are $n, 3 n, 5 n . . .$

∴ note $3 n = 255$ ⇒ $n = 85$ , note $5 n = 85 \times 5 = 425$

note $7 n = 7 \times 85 = 595$

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A student determines the velocity of sound with the help of a closed organ pipe. If the observed length for fundamental frequency is 24.7 cm, the length for third harmonic will be :

74.1 cm

72.7 cm

75.4 cm

73.1 cm

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Explanation

$l_{2} = 3 l_{1} = 3 \times 24 .7 = 74 .1 c m$

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In a resonance tube the first resonance with a tuning fork occurs at 16 cm and second at 49 cm. If the velocity of sound is 330 m/s, the frequency of tuning fork is :

500

300

330

165

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Explanation

For closed pipe $l_{1} = \frac{v}{4 n}$ ;

$l_{2} = \frac{3 v}{4 n}$ ⇒ $v = 2 n (l_{2} - l_{1})$

⇒ $n = \frac{v}{2 (l_{2} - l_{1})} = \frac{330}{2 \times (0 .49 - 0 .16)} = 500 H z$

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Two closed organ pipes of length 100 cm and 101 cm 16 beats in 20 sec. When each pipe is sounded in its fundamental mode calculate the velocity of sound

303 ms^–1

332 ms^–1

323.2 ms^–1

300 ms^–1

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Explanation

Number of beats per second,

$n = \frac{16}{20} = \frac{4}{5}$ ⇒ $n = n_{1} - n_{2} = \frac{v}{4} (\frac{1}{l_{1}} - \frac{1}{l_{2}})$

⇒ $\frac{4}{5} = \frac{v}{4} (\frac{1}{1} - \frac{1}{1 .01}) = \frac{0 .01 v}{4 \times 1 .01}$

$v = \frac{16 \times 101}{5} = 323 .2 m s^{- 1}$

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An organ pipe, open from both end produces 5 beats per second when vibrated with a source of frequency 200 Hz. The second harmonic of the same pipes produces 10 beats per second with a source of frequency 420 Hz. The frequency of source is

195 Hz

205 Hz

190 Hz

210 Hz

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Explanation

Initially number of beats per second = 5

∴ Frequency of pipe = 200 ± 5 = 195 Hz or 205 Hz ...(i)

Frequency of second harmonics of the pipe = 2n and number of beats in this case = 10

∴ 2n = 420 ± 10 ⇒ 410 Hz or 430 Hz

⇒ n = 205 Hz or 215 Hz ... (ii)

From equation (i) and (ii) it is clear that n = 205 Hz

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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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