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Q.

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 is heard. If the frequency of fork 1 is 200 Hz, then what was the original frequency of fork 2 (in Hz) ?

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answer is 196.

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Detailed Solution

Frequency of fork 1, no = 200 Hz
Number of beats heard when fork 2 is sounded with fork 1=Δn=4 
Now on loading (attaching tape) on unknown fork, the mass of tuning fork increases, So the beat frequency increase (from 4 to 6 in this case) then the frequency of the unknown fork 2 is given by.
n=n−Δn=200−4=196 Hz
 

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