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When two tuning forks are vibrated simu...

When two tuning forks are vibrated simultaneously, 4 beats are heard per second. The fork is waxed slightly and then 6 beats are heard per second when both are vibrated again simultaneously . Find out the frequency of the second fork . Given the frequency of the first fork is 510 Hz .

Text Solution

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Let `n_(2)` be the frequency of the 2nd tuning fork. We know, number of beats per second = difference in frequencies of the two tuning forks.
If ` n_(2) gt 510 Hz , then n_(2) - 510 = 4 or, n_(2) = 514 Hz` .
When the 2nd fork is waxed, its frequency becomes less than 514 Hz . then the number of beats per second becreases. So, 6 beats will not be formed per second .
If ` n_(2) lt 510 Hz , then 510 - n_(2) = 4 or, n_(2) = 506 ` Hz. The frequency decreases further for waxing . So, 6 beats can be formed per second .
This means that the value ` n_(2) = 506` Hz matches with the given problem .
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