A simple harmoinc progressive wave of amplitude 0.05 m and frequency 5Hz is travelling along the positive direction of x-axis with a speed of 40 m/s, then the displacement of a particle at 30 m from origin in the time 1 second will be

Options

A. 0.05 m
B. 0
C. 0.02 m
D. 0.025 m

Answer

A

Explanation

`y = A sin omega t - kx`
`= 0.05 sin [2pi nt - (2pi)/(lamda) x]`
`= 0.05 sin [2pi 5 xx 1 - (2pi xx 30)/((v)/(n))]`
`= 0.05 sin [10pi - (60pi xx 5)/(40)]`
`= 0.05 sin [10 pi - 7.5 pi]`
`= 0.05 sin (2.5 pi)`
`= 0.05 sin (2pi + (pi)/(2)) = 0.05 "sin" (pi)/(2)`
`= 0.05 xx 1 = 0.05 m`