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

Options

A. `(sqrt(3 pi))/(50 ) hatj` m/s
B. `- (sqrt(3 pi))/(50 ) hatj` m/s
C. `(sqrt(3 pi))/(50 ) hati` m/s
D. ` - (sqrt(3 pi))/(50 ) hati` m/s

Answer

A

Explanation

Velocity of point P = `(dy)/(dt ) = - ` (velocity of wave ) `xx (dy)/(dx) `
Here `(dy)/(dx) ` is negative . Therefore,velocity at point P is positive and is along y-axis only.
Aliter
Equations of a wave moving in positive x-a-axis is given
as y = A sin `(omega t - phi ) or V_(p) = A omega cos (omega t - phi) `
Here y = 5 cm , A = 10 cm,
`therefore 5 = 10 sin (omega t - phi) rArr omega t - phi = 30^(@)`
Subsitituting this value in the quation of velocity we get
`V_(p) = 0.10 xx omega cos 30^(@)`
Now V = v `lambda " " therefore v= (v)/(lambda) = (0.10)/(0.5) = 0.2`
`therefore omega = 2pi v = 2pi xx 0.2 = 0.4 pi`
`rArr v_(P) = 0.1 xx 0.4 pi xx (sqrt(3))/(2) = (sqrt(3))/(50 ) pi `
It has to be in positive y direction.
So correct choice is (a).