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 $200Hz$, then what was the original frequency of fork $2$?

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

Answer 1

The correct option is D $196Hz$

Frequency of fork $1=200Hz=n_0$
No. of beats heard when fork $2$ is sounded with fork $1=\Delta n=4$
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, $n=n_0-\Delta n=200-4=196Hz$