The maximum particle velocity in a wavemotion is half the wave velocity.Then the amplitude of the wave is equal to

Options

A. `lambda/(4pi)`
B. `(2lambda)/(pi)`
C. `(lambda)/(2pi)`
D. `lambda`

Answer

A

Explanation

For a wave, y = a sin `((2pivt)/(lambda) -(2pix)/(lambda))`
Here v = velocity of wave
`therefore y = asin ((2piv)/(lambda) t - (2pix)/(lambda))`
`therefore (dy)/(dt) = a((2piv)/(lambda)) cos ((2pivt)/(lambda) - (2pix)/(lambda))`
`therefore velocity = (2piav)/(lambda) cos((2pivt)/(lambda) - (2pix)/(lambda))`
`therefore` Maximum velocity is obtained when `cos ((2pivt)/(lambda) - (2pix)/(lambda))=1`
`therefore v_(m) = (2piav)/(lambda)` `therefore v_(m) = (v)/(2)`(given)
`therefore(2piav)/(lambda) = (v)/(2)` or `a = (lambda)/(4pi)`.