A 75 cm long stretched string is tuned with a tuning fork. If the length of the string is reduced by 3 cm, it produces 6 beats with the tuning fork per second. Find out the frequency of the tuning fork.

Explanation

Frequency of the tuning fork, n = fundamental frequency of the 75 cm long string ` = (1)/(2 xx 75) sqrt((T)/(m))`
Again, fundamental frequency of the (75 - 3) cm or 72 cm long string ,
`n^(.) = (1)/(2 xx 72) sqrt((T)/(m))`
`:. (n)/(n^(.)) = (72)/(75) , "clearly", n lt n^(.)` .
As 6 beats are produced per second, we have,
`n^(.) - n = 6 ` or , `(n^(.))/(n) - 1 = (6)/(n)`
or, ` (75)/(72) - 1 = (6)/(n) ` or, `(3)/(72) = (6)/(n)` or , n = 144 Hz .