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:

Options

A. 1 m
B. 75 cm
C. 60 cm
D. 50 cm

Answer

A

Explanation

As we use `lambda = 2l = 3m`
equation of standing wave
`y = 2A sin kx cos omegat`
y = A as amplitude is 2.A
A = 2A sin ks
`(2pi)/(lambda)x = (pi)/(6)`
`x_(1) = (1)/(4)m ` and `(2pi)/(lambda) .x = (pi)/(2) + (pi)/(3)`
`x_(2) = 1.25m`
`x_(2) - x_(1)` = 1m`