Question: A tuning fork gives 5 beats with another tuning fork of frequency 100 Hz. When the first tuning fork...

Question

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

Answer 1

Solution

Suppose n_A = known frequency = 100 Hz, n_B = ?

x = 5 bps, which remains unchanged after loading

Unknown tuning fork is loaded so n_B↓

Hence n_A – n_B ↓ = x ... (i)

n_B ↓ – n_A = x ... (ii)

From equation (i), it is clear that as n_B decreases, beat frequency. (i.e. n_A – (n_B)_new) can never be x again.

From equation (ii), as n_B↓, beat frequency (i.e. (n_B)_new – n_A) decreases as long as (n_B)_new remains greater than n_A, If (n_B)_new become lesser than n_A the beat frequency will increase again and will be x. Hence this is correct.

So, n_B = n_A + x = 100 + 5 = 105 Hz.