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 ?

A tuning fork A, marked 512 Hz, produces 5 beats per second, where sou

1.	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 ?
Answer» Solution :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.