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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. Minimum distance between the two points having amplitude 2 mm is:

A

`1m`

B

`75cm`

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)/(6)`
`rArr x_(1)=(1)/(4)m`
and `(2pi)/(lambda).x=(5pi)/(6)`
`rArrx_(2)=1.25m`
`rArr x_(2)-x_(1)=1m`
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