When two tuning forks (fork 1 and fork 2) are sounded simultaneou… - Vidyadip

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 $2$? .... $Hz$

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Answer

Frequency of fork $1=200 \mathrm{Hz}=\mathrm{n}_{0}$

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, $\mathrm{n}=\mathrm{n}_{0}-\Delta \mathrm{n}=200-4=196 \mathrm{Hz}$

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