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. (a)` 1 m`
B. (b)`75 cm`
C. (c)`60 cm`
D. (d)`50 cm`

Answer

A

Explanation

`lambda = 2l = 3m`
Equation of standing wave
( As `x = 0` is taken as a node)
`y = 2 A kx cos omega t`,
given ` 2 A = 4 mm`
To find value of `x` for which amplitude is `2 mm` , we have ` 2 mm = ( 4mm) sin kx`
`( 2pi)/( lambda) x = ( pi)/(6) rArr x_(1) = (1)/(4) m`
`( 2pi)/( lambda) x = (pi)/( 2) + (pi)/( 3) rArr x_(2) = 1.25 m`
`x_(2) - x_(1) = 1 m`