A progressive wave of frequency 500 HZ is travelling with a velocity of 360`ms^(-1)`. The distance between the two points, having a phase difference of `60^(@)` is ………

Options

A. 1.2 m
B. 12 m
C. 0.12 m
D. 0.012 m

Answer

C

Explanation

`f=500Hz,v=360ms^(-1)`
`lamda=v/f=360/500`
Now, a phase difference of `60^(@)` corresponds to a path difference of `60/360xxlamda`
So, distance between 2 particles is
`d=60/360xxlamda=60/360xx360/500=0.12m`