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When two tuning forks (fork 1 and fork 2...

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

A

`200 Hz`

B

`202 Hz`

C

`196 Hz`

D

`204 Hz`

Text Solution

Verified by Experts

The correct Answer is:
C
The frequency of fork `2` is `= 200 +- 4 = 196` or `204 Hz`
Since, on attaching the tape the prong of fork `2`, its frequency decrease, 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 thanthe frequency of tunin fork `1`. Hence, the frequency of fork `2 = 196 Hz`.
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