Question

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 $v$ and $f_0$ is the fundamental frequency of the pipe. For which of the following combination(s) of $u$ and $f_s$, will the sound reaching the pipe lead to a resonance?

• A. $u=0.8v$ and $f_s=f_0$
• B. $u=0.8v$ and $f_s=2f_0$
• C. $u=0.8v$ and $f_s=0.5f_0$
• D. $u=0.5v$ and $f_s=1.5f_0$

Answer 1

Answer: (A), (D)

Natural frequency of closed pipe
$f=(2n+1)f_0$
$f_0$ is fundamental frequency
$n=0,1,2,\dots \dots \dots .$
frequency of source received by pipe
$f^{\prime }=f_s\left[\frac{v-0}{v-u}\right]$
For resonance $f^{\prime }=f$
$f_s\left[\frac{v}{v-u}\right]=(2n+1)f_0$
If $u=0.8v{f}_s=f_0$
$f^{\prime }=\frac{v}{0.2v}{f}_0=5f_0$
for $n=2$ pipe can be in resonance
Hence, option (A) is correct.
If $u=0.8v{f}_s=2f_0$
$f^{\prime }=\frac{v}{0.2v}\times 2f_0=10f_0$
If $u=0.8v,f_S=0.5f_0$
$f^{\prime }=\frac{v}{0.2v}\times 0.5f_0=2.5f_0$
Not possible
If $u=0.5v,f_S=1.5f_0$
$f^{\prime }=\frac{v}{0.5v}\times 1.5f_0=3f_0$
for $n=1f=3f_0$
pipe can be in resonance
Hence, option (D) is correct.