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A sting of length 1m and linear mass den...

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,

A

`n_3 = 5n_1 = f_3 = 125 Hz`

B

`f_3 = 5f_1 = n_2 = 125 Hz`

C

`f_3 = n_2 = 3f_1 = 150 Hz `

D

`n_2 = (f_1 + f_2)/2 = 75 Hz`

Text Solution

Verified by Experts

The correct Answer is:
D
`f_1 = v/(2l) = sqrt(T//mu)/(2l)`
`= sqrt((100//0.01))/2 = 50 Hz.`
`f_2 = 2f_1 = 100Hz`
`f_3 = 3f_1 = 150 Hz`
`n_1 = v/(4l) = 25 Hz`
`n_2 = 3n_1 = 75 Hz`
`n_3 = 5n_1 = 125 Hz` .
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