A sound source (\(S\)) emits frequency of \(180~Hz\) when moving towards a rigid wall with speed \(5~m/s\) and an observer (\(O\)) is moving away from wall with speed \(5~m/s\). Both source and observer move on a straight line which is perpendicular to the wall as shown in figure. Find the number of beats per second heard by the observer. (\(\text {Speed of sound}= 355~ m/s\))
A sound source (\(S\)) emits frequency of \(180~Hz\) when moving towards a rigid wall with speed \(5~m/s\) and an observer (\(O\)) is moving away from wall with speed \(5~m/s\). Both source and observer move on a straight line which is perpendicular to the wall as shown in figure. Find the number of beats per second heard by the observer. (\(\text {Speed of sound}= 355~ m/s\))
Formula Used: \(n_1=n\left(\dfrac{\upsilon -\upsilon _0}{\upsilon -\upsilon _s} \right)\),
\(n_2=n \left(\dfrac{\upsilon -\upsilon _0}{\upsilon -\upsilon _s} \right)\)
Given, \(\upsilon = {\text {speed of sound}} = 355~m/s)\)
\(n={\text{frequency of source }}=180~Hz\)
\(n_s={\text{velocity of source}} = 5~m/s \)
\(\upsilon _0={\text{velocity of the observer}}=5~m/s\)
For direct sound, source is moving away from the observer. So, the frequency heard,
\(\Rightarrow{n_1}=n\left(\dfrac{\upsilon -\upsilon _0}{\upsilon +\upsilon _s } \right)\)
\(\Rightarrow{n_1}=180\left(\dfrac{355-5}{355+5}\right)\)
\(\Rightarrow{n_1}=180\left(\dfrac{350}{360}\right)\)
\(\Rightarrow{n_1}=175~Hz\)
The other sound is echo, reaching the observer from the wall from the image of source formed by reflection at the wall. This image is approaching the observer in the direction of sound,
Hence for reflected sound, frequency heard by the observer is,
\(\Rightarrow{n_2}=n\left(\dfrac{\upsilon -\upsilon _0}{\upsilon -\upsilon _s}\right)\)
\(\Rightarrow{n_2}=180\left(\dfrac{355-5}{355-5}\right)\)
\(\Rightarrow{n_2}=180\)
\(\text{Beat frequency}= n_2-n_1\)
\(\text{Beat frequency}= 180-175\)
\(\text {Beat frequency}= 5~Hz\)
Formula Used: \(n_1=n\left(\dfrac{\upsilon -\upsilon _0}{\upsilon -\upsilon _s} \right)\),
\(n_2=n \left(\dfrac{\upsilon -\upsilon _0}{\upsilon -\upsilon _s} \right)\)
Given, \(\upsilon = {\text {speed of sound}} = 355~m/s)\)
\(n={\text{frequency of source }}=180~Hz\)
\(n_s={\text{velocity of source}} = 5~m/s \)
\(\upsilon _0={\text{velocity of the observer}}=5~m/s\)
For direct sound, source is moving away from the observer. So, the frequency heard,
\(\Rightarrow{n_1}=n\left(\dfrac{\upsilon -\upsilon _0}{\upsilon +\upsilon _s } \right)\)
\(\Rightarrow{n_1}=180\left(\dfrac{355-5}{355+5}\right)\)
\(\Rightarrow{n_1}=180\left(\dfrac{350}{360}\right)\)
\(\Rightarrow{n_1}=175~Hz\)
The other sound is echo, reaching the observer from the wall from the image of source formed by reflection at the wall. This image is approaching the observer in the direction of sound,
Hence for reflected sound, frequency heard by the observer is,
\(\Rightarrow{n_2}=n\left(\dfrac{\upsilon -\upsilon _0}{\upsilon -\upsilon _s}\right)\)
\(\Rightarrow{n_2}=180\left(\dfrac{355-5}{355-5}\right)\)
\(\Rightarrow{n_2}=180\)
\(\text{Beat frequency}= n_2-n_1\)
\(\text{Beat frequency}= 180-175\)
\(\text {Beat frequency}= 5~Hz\)
