When two tuning forks (fork 1 and fork 2) are sounded simultaneously, 4 beats/s are heard. Now, some tape is attached on the prong of the fork2. When the tuning fork are sounded again, 6 beats/s are heard. If the frequency of fork 1 is 200Hz, then what was the original frequency of fork 2?

Question

Dr. Abubacker · Accepted Answer

(c) The frequency of fork 2
$= 200 \pm 4 = 196\ or\ 204\ Hz$
Since, on attaching the tape on the prong of fork 2, its frequency decreases, but now the number of beats per second is 6 i.e., the frequency difference now increases. It is possible only when before attaching the tape, the frequency of fork 2 is less than the frequency of tuning fork 1. Hence, the frequency of fork 2 is 196 Hz.

Q. When two tuning forks (fork 1 and fork 2) are sounded simultaneously, 4 beats/s are heard. Now, some tape is attached on the prong of the fork2. When the tuning fork are sounded again, 6 beats/s are heard. If the frequency of fork 1 is 200Hz, then what was the original frequency of fork 2?

Explanation

Reviewed by Dr. Abubacker

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