Questions
A tuning fork gives 5 beats with another tuning fork of frequency 100 Hz. When the first tuning fork is loaded with wax, then the number of beats remains unchanged, then what will be the frequency of the first tuning fork
detailed solution
Correct option is C
Suppose nA = known frequency = 100 Hz, nB = ? x = 5 bps, which remains unchanged after loadingUnknown tuning fork is loaded so nB↓Hence nA – nB ↓ = x ... (i) nB ↓ – nA = x ... (ii) From equation (i), it is clear that as nB decreases, beat frequency. (i.e. nA – (nB)new) can never be x again.From equation (ii), as nB↓, beat frequency (i.e. (nB)new – nA) decreases as long as (nB)new remains greater than nA, If (nB)new become lesser than nA the beat frequency will increase again and will be x. Hence this is correct.So, nB = nA + x = 100 + 5 = 105 Hz.Similar Questions
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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