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When tuning forks A and B are sounded to...

When tuning forks A and B are sounded together, 20 beats are produced in 8 second between them. (i.e. beat frequency is `(20)/(8) =2.5 Hz).` Now, when some wax is applied on one of these tuning forks, 32 beats are produced in 8 s. If frequency of that fork on which wax is not applied is 512 Hz then find frequency of another tuning fork, before and after applying wax.

Text Solution

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Suppose wax is not applied on tuning for A. Hence, as per the statement, we can take its frequency `f _(A) =512 Hz`
Now if frequency of B before applying wax is `f _(B)` then,
`f _(A) = 512 Hz {{:( f _(B) =512 + 2.5= 514.5 Hz), (f _(B) = 512-2.5 = 509.5 Hz):}`
`f ._(B) = 508 Hz`
`(because f ._(B) lt f _(B) and f _(A) - f ._(B) = 4 Hz)`
After applying wax on B, its frequency becaomes `f ._(B), ` where `f ._(B) lt f _(B)` because as mass increases, frequency decreases.
Now, new beat frequency is `(32)/(0) =4 Hz`
`implies f _(B) = 514.5 Hz OR f _(B) = 509.5 Hz`
and `f ._(B) = 508 Hz` such that:
`(i)f _(A) -f_(B)= 514.5 - 512=512 -509.5 = 2.5 Hz`
(ii)` f ._(B) lt f _(B)`
(iii) `f _(A) -f ._(B) = 512 -508 =4 Hz`
Before applying wax on `B, f _(B) =514 .5 Hz OR 509.5 Hz and ` after applying wax on B, `f ._(B) = 508 Hz.`
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