A transverse sinusoidal wave moves along a string in the positive x-direction at a speed of `10cm//s`. The wavelength of the wave is 0.5 m and its amplitude is `10 cm`. At a particular time t,the snapshot of the wave is shown in figure. The velocity of point P when its displacement is 5 cm is

(a) `sqrt(3pi)/(50)hatjm//s`
(b)`-sqrt(3pi)/(50)hatjm//s`
(c)`sqrt(3pi)/(50)hatim//s`
(d) `-sqrt(3pi)/(50)hatim//s`

Options

A. `(sqrt(3) pi)/(50) hat(j) m//s`
B. `- (sqrt(3) pi)/(50) hat (j) m//s`
C. `(sqrt(3) pi)/(50) hat(j) m//s`
D. `- (sqrt(3) pi)/(50) hat(i) m//s`

Answer

A

Explanation

Take `y = A sin (omega t - kx)`
so `V_(p) = (del y)/(del t)`
`=- A cos omega cos(omega t - kx)`
`rArr V_(P)=omegasqrt(A^(2)-y^(2))=((2piv)/(lamda))sqrt(A^(2)-y^(2))`
`=(2pi(10xx10^(-2)))/(0.5)(sqrt((10)^(2)-(5)^(2)))xx10^(-2)`
`=(sqrt(3)pi)/(50)ms^(-1)`