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
Suppose nA = known frequency = 100 Hz, nB = ?
x = 5 bps, which remains unchanged after loading
Unknown 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.
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
