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A string of length 1.5 m with its two en...

A string of length 1.5 m with its two ends clamped is vibrating in fundamental mode Amplitude at the centre of the string is 4 mm. Distance between the two points having amplitude 2mm is

A

1m

B

75 cm

C

60 cm

D

50 cm

Text Solution

Verified by Experts

The correct Answer is:
A
`lambda=2l=3m`
Equation of standing wave
`y=2A sin kx cos omegat`
y=A as amplitude is 2A.
`A=2A sin kx`
`(2pi)/(lambda)x =(pi)/6rArr x_(1)=1/4m`
and `(2pi)/(lambda)x=(5pi)/6rArr x_(2)=1.25 m rArr x_(2)-x_(1)=1m`
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