A tuning fork is found to give ` 20 beats "in" 12 s` when sounded in conjuction with a stretched string vibrating under a tension of `10.2 or 9.9 kgf`. Calculate the frequency of the fork.

Explanation

` n = (1)/(2 I) sqrt((T)/(m))`
`:. N_(1) = (1)/( 2I) sqrt (( 10.2 g)/(m)) and n_(2) = (1)/( 2 I) sqrt (( 9.9 g)/( m))`
Dividing `(n_(1))/( n_(2)) = sqrt (( 10.2)/(9.9)) ( Obviously n_(1) gt n_(2))`
`:. N_(1) - N = ( 20)/( 12) and N - n_(2) = (20)/(12)`
or `(n_(1))/( n_(2)) = (N + (20)/(12))/(N - (20)/(12))`
`:. (N + (20)/(12))/(N - (20)/(12)) = sqrt ((10.2)/(9.9)) or N = 223 Hz`