Consider a sinusoidal travelling wave shown in figure. The wave velocity is `+ 40 cm//s`.
Find
(a) the frequency
(b) the phase difference between points `2.5 cm` apart
(c) the velocity of a particle at `P` at the instant shown.

Answer

A, B, C, D

Explanation

(a) `(lambda)/(2) = 2 cm`
rArr `lambda=4 cm`
`f =(v)/(lambda) = (40)/(4) =10 Hz`
(b) `Deltaphi = (2pi)/(lambda)Deltax = ((2pi)/(4))(2.5) = (5pi)/(4)`
( c ) `omegat = theta`
or `(2pi//t) = theta`
`:. t=(theta)/(2pif) = (pi//3)/((2pi)(10))`
`=(1)/(60)s`
(d) At `P`, particle is at mean position. So,
`v= "maximum velocity"`
`= omegaA`
`=2pifA`
`=(2pi)(10)(2)`
`= (40pi) cm//s`
`=125.7 cm//s = 1.26 m//s`
Further, `v_(P) = - (v)((dely)/(delx))` ...(i)
Sign of `v_(1)`, the wave velocity is given positive.
Sign of `(dely)/(delx)`, slope of `y - x` graph is also positive.
Hence, from Eq. (i) particle velocity is negative.
`:. v_(P) = - 1.26 m//s`