The frequency of tuning forks `A` and `B` are respectively `3%` more and `2%` less than the frequency of tuning fork `C`. When `A` and `B` are simultaneously excited, 5 beats per second are produced. Then the frequency of the tuning fork `A` (in Hz) Is

Options

A. `98`
B. `100`
C. `103`
D. `105`

Answer

C

Explanation

Let `n` the frequency of fork `C` then
`n_A=n+(3n)/(100)=(103n)/(100)` and `n_B=n-(2n)/(100)=(98)/(100)`
but `n_A-n_B=5implies(5n)/(100)=5impliesn=100Hz`
`:. n_A=((103)(100))/(100)=103Hz`