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 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. $204\,Hz$

Answer 1

Answer: (C)

$196\,Hz$

Answer 2

The frequency of fork 2 $=200\pm4=196 or 204\,Hz$ 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 $196\, Hz$.