A `10kg` weight is suspended with a wire `AB` of linear mass density `0.01kg//mj`. Length of `AP` segment is `1m` and it is vibrated with a tuning form in its second overtone. Now instead of `10kg`. A `40kg` weight is suspended. By what minimum distance should we displace the pulley forward to achieves the resonance with same tuning fork `(g=10m//sec^(2))`

Options

A. `0.33m`
B. `0.67m`
C. `1m`
D. `1.33m`

Answer

A

Explanation

Initially
`f_(0)=3/(2l)sqrt(T/(mu))`
`t_(0)=3/2 3/(2xx1)sqrt(100/0.01)=150Hz`
In the second case suppose the tunning fork is in resonance with `n^(th)` harmonic
`f_(0)=n/(3l)sqrt(T/(mu))`
`150=n/(2l)sqrt(400/0.01)` `l=n/1.5`
`n=1 l_(1)=1/1.5=0.67`
`n=2 l_(2)2/1.5=1.33`
`n=3 l_(3)=3/1.5=2m`