# 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 th

19 views
in Physics
closed
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. 200 Hz
B. 202 Hz
C. 196 Hz
D. 204 Hz

by (85.2k points)
selected

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.