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. $u =0.8 \,v$ and $f _{ s }= f _{0}$
• B. $u=0.8\, v$ and $f _{ s }=2\, f _{0}$
• C. $u =0.8 \,v$ and $f _{ s }=0.5\, f _{0}$
• D. $u=0.5 \,v$ and $f _{ s }=1.5 \,f _{0}$

Answer 1

Answer: (A), (D)

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