When 2 tuning forks (fork 1 and fork 2) are sounded simultaneously, 4 beats $s^{- 1}$ are heard. Now, some tape is attached on the prong of fork 2. When the tuning forks are sounded again, 6 beats $s^{- 1}$ are heard if the frequency of fork 1 is 200 Hz, then what was the original frequency of fork 2?

Question

Dr. Abubacker · Accepted Answer

(a) Here$n_{1} = 200\ Hz$.
Number of beats $s^{- 1};m = 4$
$	herefore\ n_{2} = 200 \pm 4 = 204\ or\ 196\ Hz$
On loading fork 2, its frequency decreases. And number of beats $s^{- 1}$ increases to 6. Therefore $m$ is negative.
$n_{2} = 200 - 4 = 196\ Hz$

Explanation

Reviewed by Dr. Abubacker

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