A sting of length 1m and linear mass density 0.01 `kg//m` is stretched to a tension of 100 N. When both ends of the string are fixed, the three lowest frequencies for standing wave are `f_1, f_2 and f_3` . When only one end of the string is fixed, the three lowest frequencies for standing wave are `n_1, n_2 and n_3`. Then,

Options

A. `n_(3)=5n_(1)=f_(3)=125Hz`
B. `f_(3)=5f_(1)=n_(2)=125Hz`
C. `f_(3)=2n_(2)=3f_(1)=150Hz`
D. `3f_(1)=n_(2)=f_(3)=75Hz`

Answer

D

Explanation

`f_(0)=(1)/(2l)sqrt((T)/(mu))=(1)/(2)sqrt((100)/(0.01))=50Hz implies f_(1),f_(2),f_(3)` are respectively 50 Hz, 100 Hz and 150 Hz
`n_(0)=(1)/(4l)sqrt((T)/(mu))=25Hz implies n_(1),n_(2) & n_(3)` are respectively 25Hz, 75Hz & 125 Hz respectively.