Question

When two tuning forks (fork `1` and fork `2`) are sounded simultaneously, `4` beats per second are heard. Now some tape is attached on the prong of the fork `2`. When the tuning fork 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. (a) `202 Hz`
B. (b) `200 Hz`
C. ( c) `204 Hz`
D. (d) `196 Hz`

Answer 1

Correct Answer - D
(d) No. of beats heard when fork `2` is sounded with fork `1 = Deltan = 4`
Now we know that if on loading (attaching tape) an unknown fork, the beat frequency increases (from `4 to 6` in this case) then the frequency of the unknown fork `2` is given by,
`n = n_(0) - Deltan = 200 - 4 = 196 Hz`