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 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` ?

Options

A. 200 Hz
B. 202 Hz
C. 196 Hz
D. 204 Hz

Answer

C

Explanation

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.