Question

A tuning fork of frequency 500 Hz gives beats of 5 Hz with another tuning fork. frequency remains unchanged. Frequency of the other fork before waxing is​

Answer 1

Answer:

Correct option is

C

105Hz

Suppose n= known frequency =100Hz,n=?

x=5bps, which remains unchanged after loading

Unknown tuning fork is loaded so n↓

Hence n−n↓=x ...(i)

n↓−n=x ...(ii)

From equation (i), it is clear that as n B

decreases, beat

frequency. (i.e.n−(n)) can never be x again.

From equation (ii), as n↓, beat frequency (i.e.(n)

–

n) decreases as long as (n)_ remains greater than n

1

, If (n)_ become lesser

than n the beat frequency will increase again and will be x.Hence this is correct.

So n = n + x = 100 + 5 = 105 Hz

Explanation:

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