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Waves

Section: Physics

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Question : 7 of 31

Marks: +1, -0

A source, approaching with speed

u

towards the open end of a stationary pipe of length

L

, is emitting a sound of frequency

f_{s^{-}}

. The farther end of the pipe is closed. The speed of sound in air is

v

f_0

is the fundamental frequency of the pipe. For which of the following combination(s) of

u

f_s

, will the sound reaching the pipe lead to a resonance ?

[JEE Adv 2021 P2]

$u=0.8 v$ and $f_s=f_0$
$u=0.8 v$ and $f_s=2 f_0$
$u=0.8 v$ and $f_s=0.5 f_0$
$u=0.5 v$ and $f_s=1.5 f_0$

Solution:

From Doppler's effect

f=f_s\left(\;\frac{v}{v-u}\right)

As the pipe is closed at one end

\therefore

For resonance,

f=f_0\left(\;\frac{v}{v-u}\right)=n f_0

where

n=

odd integer

(a) For,

u=0.8 v

and

f_s=f_0

,

f=f_0\left(\;\frac{v}{v-0.8 v}\right)=5 f_0

(b) For,

u=0.8 v

and

f_s=2 f_0

,

f=2 f_0\left(\;\frac{v}{v-0.8 v}\right)=10 f_0

(c) For,

u=0.8 v

and

f_s=0.5 f_0

f=0.5 f_0\left(\;\frac{v}{v-0.8 v}\right)=2.5 f_0

(d) For

u=0.5 v

and

f_s=1.5 f_0

f=1.5 f_0\left(\;\frac{v}{v-0.5 v}\right)=3 f_0

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