The tension of a stretched string is increased by 69%. In order to keep its frequency of vibration constant, its length must be increased by

Options

A. `30%`
B. `20%`
C. `69%`
D. `sqrt(69)%`

Answer

A

Explanation

Frequency of vibration of a stretched string is given by: `n=(1)/(2l)sqrt((T)/(m))`
Tension is increased by 69%.
`therefore`New tension, `T.=T+(69)/(100)T=(169T)/(100)`
Suppose to keep the frequency same, the length of the string is to be increased by x%.
`therefore`New length`=l.=l+(lx)/(100)=((100+x)/(100))l`
`therefore`Changed frequency`=n.=(1)/((2(100+x)l)/(100))sqrt((169T)/(100m))`
Now, by problem, `n.=n`.
`therefore(1)/((2(100+x)l)/(100))sqrt((169T)/(100m))=(1)/(2l)sqrt((T)/(m))`
`therefore (100)/(100+x)*(13)/(10)=1impliesx=30`.