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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» <html><body><p></p>Solution :Here `<a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a> _(A) = 512 Hz implies f _(B) = <a href="https://interviewquestions.tuteehub.com/tag/limlimits-y-rightarrow-2-fracy2-4y-2-517-324511" style="font-weight:bold;" target="_blank" title="Click to know more about 517">517</a> Hz or 507 Hz` <br/> `(because ` Initial beat <a href="https://interviewquestions.tuteehub.com/tag/frequency-465761" style="font-weight:bold;" target="_blank" title="Click to know more about FREQUENCY">FREQUENCY</a> between A and B is <a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a> Hz) <br/> Now, when some wax is applied on B, its frequency will decrease and will assume that value `f._(B)` such that, <br/> `(i) f ._(B) lt f _(B.) (ii) f _(A)~f._(B) = 5 Hz` <br/> `(because` Final beat frequency beetween A and B is also 5 Hz) <br/> `implies f_(B)=517 Hz and f._(B) = 507 Hz` <br/> because `507 lt 517 impliesf._(B) lt f _(B)` <br/> `and f _(B) -f _(A) =517 -512 =5 Hz` <br/> and `f _(A) -f _(B) = 512 -507 = 5 Hz` <br/> 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.</body></html>


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