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`

A string of length 1.5 m with its two ends clamped is vibrating in fun

1.	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`
Answer» Correct Answer - A `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`

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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`

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