The equations of a transverse wave is given by `y= 0.05 sin pi (2t- 0.02x)`, where x,y are in metre and t is in second. The minimum distance of separation between two particles which are in phase and the wave velocity are respectively…….

Options

A. `50m, 50 ms^(-1)`
B. `100m, 100 ms^(-1)`
C. `50m, 100 ms^(-1)`
D. `100m, 50 ms^(-1)`

Answer

B

Explanation

The given equation of a transverse wave is `y = 0.05 sinpi(2t - 0.02x) = 0.05 sin(2pi t - 0.02 pi x)`
Compare it with the standard equation,
`y = A sin(omega t - kx)`
We get
A = 0.05 m, `omega = 2 pi rad s^(-1)`
`k = 0.02 pi rad m^(-1)`
The minimum distance between points having the same phase is known as wavelength of the wave,
`lamda = (2pi)/(k) = (2pi)/(0.02pi)` = 100m
Wave velocity, `v = (omega)/(k) = (2pi rad s^(-1))/(0.02 pi rad m^(-1)) = 100 m s^(-1)`