A metal wire of diameter $$1.5$$ mm is held on two knife edges separated by a distance $$50$$ cm the tension in the wire is $$100$$ N the wire vibrating with its fundamental frequency and vibrating tuning fork together produces $$5$$ beats per second. The tension in the wire is then reduced to $$81$$ N, when the two are excited, beats are heard at the same rate. Calculate frequency of the fork.

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Infinity Learn · Accepted Answer

Let N be the frequency of the tuning fork then the frequency of wire, when tension is $$100$$ N will be (N+5) and the tension is $$81$$ N it is (N-5) since each case $$5$$ beats are heard per second.∴$$N+5=12lT1$$μ$$=12$$×$$0.5100$$μ$$=10$$μ→(i)  and  N−$$5=12lT2$$μ$$=12$$×$$0.581$$μ$$=9$$μ→(ii) Subtracting (ii) $$from(i)$$ we have $$10=1$$μ   (or)  μ$$=0.01$$ $$kg/m$$   Using this value in (i) (or) (ii) $$N=95$$​ Hz

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