The stirng stretched by tension T and length L vibrates in resonance with a tuning fork of frequency n. the tension in the stretched string is increased by 69% and length of string reduced by 35%. Then, the frequency of vibrating string is

Options

A. n
B. `1.5n`
C. `2n`
D. `(n)/(2)`

Answer

C

Explanation

The fundamental frequency off vibration of the stretched string as
`n=(1)/(2L)sqrt((T)/(m))` . . . (i)
where, m is the mass per unit length of the string.
When the tension in the string is increased by 69% and length is reduced by 35%, the new tension and length become
`T'=T+(69)/(100)T=(169)/(100)T`
and `L'=L-(35)/(100)L=(65)/(100)L`
the new frequency of vibrating string is
`n'=(1)/(2L')sqrt((T')/(m))`
`=(1)/(2((65)/(100)L))sqrt(((169)/(100)T)/(m))`
`=((1300)/(650))((1)/(2L)sqrt((T)/(m)))`
`=2n` [unit eq. (i)].