KCET · Physics · Waves and Sound
When two tuning forks \( A \) and \( B \) are sounded together. \( 4 \) beats per second are heard. The
frequency of the fork \( \mathrm{B} \) is \( 384 \mathrm{~Hz} \). When one of the prongs of the fork \( \mathrm{A} \) is filed and sounded
with \( B \), the beat frequency increases, then the frequency of the fork \( A \) is
- A \( 380 \mathrm{~Hz} \)
- B \( 388 \mathrm{~Hz} \)
- C \( 379 \mathrm{~Hz} \)
- D \( 389 \mathrm{~Hz} \)
Answer & Solution
Correct Answer
(B) \( 388 \mathrm{~Hz} \)
Step-by-step Solution
Detailed explanation
Given, beat frequency \(=4\) beats per second; frequency of fork \(B=384 \mathrm{~Hz}\)
Now, we know beat frequency \(f_{A}-f_{B}\)
Here \(f_{A}\) is frequency of fork \(A ; f_{B}\) is frequency of fork \(B\)
\(\Rightarrow 4=f_{A}-384 \Rightarrow f_{A}=4+384=388\)
Therefore, frequency of the fork \(\mathrm{A}\) is \(388 \mathrm{~Hz}\).
Now, we know beat frequency \(f_{A}-f_{B}\)
Here \(f_{A}\) is frequency of fork \(A ; f_{B}\) is frequency of fork \(B\)
\(\Rightarrow 4=f_{A}-384 \Rightarrow f_{A}=4+384=388\)
Therefore, frequency of the fork \(\mathrm{A}\) is \(388 \mathrm{~Hz}\).
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