A transverse sinusoidal wave moves 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 ampli- tude 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 Figure :

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(i) m//s`
D. `- (sqrt(3) pi)/(50) hat(i) m//s`

Answer

A

Explanation

This is transverse wave travelling along positive X-axis hence particles are moving along Y-axis. Since wave is travelling in positive direction hence particle P will follow the motion of particles to its left. Particle on the left side is above point P hence point P is moving upward at this instant. So velocity of P is along positive y-direction. We can see that option (a) should be correct and we can also find its magnitude.
Amplitude (A) = 10 cm
Wavelength `(lambda)` = 50 cm
Speed of wave `(upsilon)` = 10 cm/s
`upsilon = v lambda rArr v = (10)/(50) = 0.2` Hz
`omega = 2 pi v = 0.4 pi`
Speed of the particle in terms of displacement can be written as follows :
`upsilon = omega sqrt(A^(2) - y^(2))`
`upsilon = 0.4 pi [ sqrt((10)^(2) - (5)^(2))] = 0.4 pi sqrt(75) = 2pi sqrt(3) cm//s`
`upsilon = (2pi sqrt(3))/(100) = (pi sqrt(3))/(50) m//s`
We can see that magnitude matches with option (a) and we already know that particle is moving along positive y-direction at the given instant. Hence option (a) is correct.