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A tuning fork A, marked 512 Hz, produces...

A tuning fork A, marked 512 Hz, produces 5 beats per second, where sounded with another unmarked tuning fork B. If B is loaded with wax the number of beats is again 5 per second. What is the frequency of the tuning fork B when not loaded ?

Text Solution

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Here `f _(A) = 512 Hz implies f _(B) = 517 Hz or 507 Hz`
`(because ` Initial beat frequency between A and B is 5 Hz)
Now, when some wax is applied on B, its frequency will decrease and will assume that value `f._(B)` such that,
`(i) f ._(B) lt f _(B.) (ii) f _(A)~f._(B) = 5 Hz`
`(because` Final beat frequency beetween A and B is also 5 Hz)
`implies f_(B) =517 Hz and f._(B) = 507 Hz`
because `507 lt 517 impliesf._(B) lt f _(B)`
`and f _(B) -f _(A) =517 -512 =5 Hz`
and `f _(A) -f _(B) = 512 -507 = 5 Hz`
Note: Here `f _(B) ne 507Hz` because otherwise after applying wax on `B, f ._(B) lt 507Hz` and then `f _(A) -f._(B) gt 5 Hz` which would be wrong as per the data given in the statement.
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