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 forks are sounded again, 6 beats per seconds 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

Answer 1

Correct Option (b) 202 Hz

Explanation:

The frequency of fork 2 = 200 ± 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.