KCET · Physics · Waves and Sound
A person with vibrating tuning fork of frequency \(338 \mathrm{~Hz}\) is moving towards a vertical wall with a speed of \(2 \mathrm{~ms}^{-1}\). Velocity of sound in air is \(340 \mathrm{~ms}^{-1}\). The number of beats heard by that person per second is
- A 2
- B 4
- C 6
- D 8
Answer & Solution
Correct Answer
(B) 4
Step-by-step Solution
Detailed explanation
As the person having tuning fork is moving towards a wall, therefore,
\(v_{s}=v_{0}=2 \mathrm{~m} / \mathrm{s} \text { (given) } \)
\( \text {So, } \frac{f^{\prime}}{f}=\frac{v+v_{0}}{v-v_{0}} \)
\( \Rightarrow \frac{f^{\prime}-f}{f}=\frac{v+v_{0}-v+v_{0}}{v-v_{0}} \)
\( \Rightarrow \frac{\Delta f}{f}=\frac{2 v_{0}}{v-v_{0}} \)
\( \text {or } \Delta f=\frac{2 v_{0}}{v-v_{0}} \times f=\frac{2 \times 2 \times 338}{(340-2)}=4\)
\(v_{s}=v_{0}=2 \mathrm{~m} / \mathrm{s} \text { (given) } \)
\( \text {So, } \frac{f^{\prime}}{f}=\frac{v+v_{0}}{v-v_{0}} \)
\( \Rightarrow \frac{f^{\prime}-f}{f}=\frac{v+v_{0}-v+v_{0}}{v-v_{0}} \)
\( \Rightarrow \frac{\Delta f}{f}=\frac{2 v_{0}}{v-v_{0}} \)
\( \text {or } \Delta f=\frac{2 v_{0}}{v-v_{0}} \times f=\frac{2 \times 2 \times 338}{(340-2)}=4\)
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