Gujarati
14.Waves and Sound
medium

Vibrating tuning fork of frequency $n$ is placed near the open end of a long cylindrical tube. The tube has a side opening and is fitted with a movable reflecting piston. As the piston is moved through $8.75 cm$, the intensity of sound changes from a maximum to minimum. If the speed of sound is $350 \,m/s. $ Then $n$ is .... $Hz$

A

$500$

B

$1000 $

C

$2000$

D

$4000$

Solution

(b) When the piston is moved through a distance of $8.75cm$, the path difference produced is $2 \times 8.75\,cm = 17.5\,cm$. This must be equal to$\frac{\lambda }{2}$ for maximum to change to minimum.

$\therefore$ $\frac{\lambda }{2} = 17.5 cm$

==> $\lambda = 35cm = 0.35m$

So, $v = n\lambda $==> $n = \frac{v}{\lambda } = \frac{{350}}{{0.35}} = 1000Hz$

Standard 11
Physics

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