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Q.

A stretched string of 1 m length, fixed at both ends, having amass of 5 x 10-4 kg is under tension of 20 N It is plucked at a point situated at 25 cm from one end. The stretchered string would vibrate with a frequency of

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a

400Hz

b

100 Hz

c

200 Hz

d

256 Hz

answer is C.

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Detailed Solution

n=p2I(Tm) Fundamental frequency n, is given by n1=12×1[205×10−4]1/2=100Hz When the wire is plucked at 25 cm, it will vibrate in two segments ∴  Frequency n2=2n1=200Hz
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