
A sound source is moving with speed \[5{m}/{s}\;\] towards a wall. If the velocity of sound is \[330{m}/{s}\;\], then the stationary observer would hear beats equal to (frequency of source = 240 Hz):
A. 0
B. 9
C. 8
D. 4
E. None of the above
Answer
514.8k+ views
Hint: The given problem is a special case of the Doppler effect. The formula that relates the frequencies of the observer, source and the velocity of the sound should be used to solve this problem. Then, to find the number of beats, we will subtract the frequency of source and observer. The number of beats equals the beat frequency.
Formula used:
\[{{f}_{0}}={{f}_{s}}\left[ \dfrac{v-{{v}_{0}}}{v-{{v}_{s}}} \right]\]
Complete answer:
From the data, we have the data as follows.
A sound source is moving with speed\[5{m}/{s}\;\]towards a wall.
\[{{v}_{s}}=5{m}/{s}\;\]
The speed of the observer will be opposite to that of the sound source.
\[\begin{align}
& {{v}_{0}}=-{{v}_{s}} \\
& \Rightarrow {{v}_{0}}=-5{m}/{s}\; \\
\end{align}\]
The velocity of sound is\[330{m}/{s}\;\].
\[v=330{m}/{s}\;\]
The frequency of the source is 240 Hz.
\[{{f}_{s}}=240Hz\]
The formula used is:
\[{{f}_{0}}={{f}_{s}}\left[ \dfrac{v-{{v}_{0}}}{v-{{v}_{s}}} \right]\]
Where \[{{f}_{0}}\] is the observer frequency, \[{{f}_{0}}\]is the source frequency, v is the velocity of the sound and \[{{v}_{0}}\] is the velocity of the observer.
Substitute the given values in the above equation to find the value of the frequency of the stationary observer.
\[\begin{align}
& {{f}_{0}}=240\left[ \dfrac{330-(-5)}{330+5} \right] \\
& \Rightarrow {{f}_{0}}=247Hz \\
\end{align}\]
Now we will compute the beat frequency.
The beat frequency is equal to the number of beats the observer hears and is given as \[{{f}_{s}}-{{f}_{0}}\].
\[\begin{align}
& {{f}_{s}}-{{f}_{0}}=247-240 \\
& \Rightarrow {{f}_{s}}-{{f}_{0}}=7 \\
\end{align}\]
As the value of the beat frequency, the stationary observer would hear beats equal to 7.
So, the correct answer is “Option E”.
Note:
The formulae are different for the different cases, that is, when the observer moves away from the source, then, the velocity of the observer should be subtracted and when the observer moves towards the source, then, the velocity of the observer should be added. The units of the parameters should be taken care of.
Formula used:
\[{{f}_{0}}={{f}_{s}}\left[ \dfrac{v-{{v}_{0}}}{v-{{v}_{s}}} \right]\]
Complete answer:
From the data, we have the data as follows.
A sound source is moving with speed\[5{m}/{s}\;\]towards a wall.
\[{{v}_{s}}=5{m}/{s}\;\]
The speed of the observer will be opposite to that of the sound source.
\[\begin{align}
& {{v}_{0}}=-{{v}_{s}} \\
& \Rightarrow {{v}_{0}}=-5{m}/{s}\; \\
\end{align}\]
The velocity of sound is\[330{m}/{s}\;\].
\[v=330{m}/{s}\;\]
The frequency of the source is 240 Hz.
\[{{f}_{s}}=240Hz\]
The formula used is:
\[{{f}_{0}}={{f}_{s}}\left[ \dfrac{v-{{v}_{0}}}{v-{{v}_{s}}} \right]\]
Where \[{{f}_{0}}\] is the observer frequency, \[{{f}_{0}}\]is the source frequency, v is the velocity of the sound and \[{{v}_{0}}\] is the velocity of the observer.
Substitute the given values in the above equation to find the value of the frequency of the stationary observer.
\[\begin{align}
& {{f}_{0}}=240\left[ \dfrac{330-(-5)}{330+5} \right] \\
& \Rightarrow {{f}_{0}}=247Hz \\
\end{align}\]
Now we will compute the beat frequency.
The beat frequency is equal to the number of beats the observer hears and is given as \[{{f}_{s}}-{{f}_{0}}\].
\[\begin{align}
& {{f}_{s}}-{{f}_{0}}=247-240 \\
& \Rightarrow {{f}_{s}}-{{f}_{0}}=7 \\
\end{align}\]
As the value of the beat frequency, the stationary observer would hear beats equal to 7.
So, the correct answer is “Option E”.
Note:
The formulae are different for the different cases, that is, when the observer moves away from the source, then, the velocity of the observer should be subtracted and when the observer moves towards the source, then, the velocity of the observer should be added. The units of the parameters should be taken care of.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

