A tuning fork of frequency n is sounded at the open end of a long cylindrical tube having a side opening and fitted with a movable reflecting piston figure. On moving the piston through 9 cm, the intensity of sound heard by the listener changes from max, to minimum. If speed of sound is 360 m/s, value of n is

1000 hz

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250 hz

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500 hz

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750 hz

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Solution

The correct option is A 1000 hz
$W h e n P i s t o n m o v e a d i s t a n c e x_{1}, P a t h d i f f e r e n c e c h a n g e b y 2 x$

$T h e r e f o r e, t h e p a t h d i f f e r e n c e b e t w e e n m a x i m a a n d c o n s e c u t i v e m i n i m a =$
$2 x = \frac{λ}{2}$
0r
$λ = 4 x = 4 \times 9 c m = 36 c m = 0.36 m$

$n = \frac{v}{λ} = \frac{360}{0.36} = 1000 H z$

$H e n c e, t h e f r e q u e n c y i s 1000 H z$

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The correct option is A 1000 hzWhen Piston move a distance x1, Path difference change by 2xTherefore, the path difference between maxima and consecutive minima=2x=λ2 0r λ = 4x = 4× 9cm = 36cm=0.36m n=vλ=3600.36=1000HzHence, the frequency is 1000Hz