A plane progressive wave of frequency 25 Hz, amplitude `2.5 xx 10^(-5) m` and initial phase zero moves along the negative x-direction with a velocity of 300 m/s. A and B are two points 6 m apart on the line of propagation of the wave. At any instant the phase different between A and B is `phi`. The maximum difference in the displacements of particle at A and B is `Delta`

Answer

5

Explanation

`v=25Hz`
`A=2.5xx10^(-5)m`
`V=300m//s`
We know that `v=vlamda" "therefore lamda=(V)/(v)=(300)/(25)=12m`
Phase difference`=phi=(2pi)/(lamda)xx`(Path difference)
`=(2pi)/(12)xx6=pi` radian
Maximum difference in displacement is equal to twice its amplitude i.e., `5xx10^(-5)m`