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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

1 m

B

75 cm

C

60 cm

D

50 cm

Text Solution

Verified by Experts

The correct Answer is:
A
`lamda = 2l = 3m`
Equation of standing wave
`y = 2A sin kx cos omega t`
`y = A` as amplitude is 2A
A = 2 A sin kx
`(2pi)/(lamda) xx = (pi)/(6) rArr x_(1) = (1)/(4) m`
and `(2pi)/(lamda) xx = (5pi)/(6)`
`rArr x_(2)=1.25 m`
`rArr x_(2)\-x_(1)=1m`
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