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)hatjm//s`
B. `-(sqrt(3)pi)/(50)hatjm//s`
C. `(sqrt(3)pi)/(5)hatim//s`
D. `-(sqrt(3)pi)/(50)hatim//s`

Answer

A

Explanation

Particle velocity `v_p=-v("slope of y-x graph ")`
Here v= -ve, as the wave is travelling in possitive direction. Slope at P is negative.
Therefore ,velocity of particle is in possitive y (or j) direction, `v=omega sqrt(A^(2)-x^(2))`
Where `omega =KC=((2pi)/(lambda))c=0.4 pi,` so,`v=0.4pi sqrt(((10^(2)/(100))-(5^(2)/(100))^(2))`