A string under a tension of 129.6 N produces 10 beats persecond when it is vibrated along with a tuning fork. When the tension in the string is increased to 160 N, it sounds in unison with the same tuning fork. Calculate the fundamental frequency oftuning fork. .

Explanation

Let n be the frequency of turning fork. The frequency of string will be either (n+10) or (n - 10) As the tension in the string increase, its frequency increases `(n infty sqrt(T))` and becomes n. This shows that the initial frequency (at = T = 129.6N) of string will be (n - 10). Hence
`n - 10 = (1)/(2l) sqrt(((129.6)/(mu)))`
`n = (1)/(2l) sqrt(((160)/(mu)))`
Dividingequation-(6.246)by equation-(6.247),weg
`(n - 10)/(n) = sqrt(((129.6)/(160)))`
Solving we get n = 100 Hz.