The figure shows a snap photograph of a vibrating string at `t = 0`. The particle `P` is observed moving up with velocity `20sqrt(3) cm//s`. The tangent at `P` makes an angle `60^(@)` with x-axis.

(a) Find the direction in which the wave is moving.
(b) Write the equation of the wave.
(c) The total energy carries by the wave per cycle of the string. Assuming that the mass per unit length of the string is `50g//m`.

Answer

A, B, C, D

Explanation

(a) `v_(P) = - v((dy)/(dx))`
As `v_(P) `and `("slope")_(P)` are both positive, `v` must be negative. Hence the wave is moving in negative x-axis.
(b) `y = A sin (omegat - kx + phi)` …(i)
`k=(2pi)/(lambda) = (pi)/(2) cm^(-1)`
`A = 4 xx 10^(-3)m =0.4 cm`
At `t =0,x = 0 "slope" (dy)/(dx) = +ve`
`:. v_(P) = - v("slope")= +ve`
`:. v_( p) = -v("slope")=+ve`
Further at `t = 0, x=0, y =+ve`
`:. phi=(pi)/(4)`
Further, `20sqrt(3) = - v tan60^(@)`
`:. v =-20 cm//s`
`f = (v)/(lambda) =5 Hz`
`:. omega = 2pif = 10pi`
`:. y = (0.4 cm) sin (10pit +(pi)/(2)x + (pi)/(4))`
( c ) `P =2pi^(2) A^(2)f^(2)muv`
`:.` Energy carried per cycle
`E = PT = (P)/(f) = 2pi^(2)A^(2)fmuv`
Subsitituting he values, we have
`E =1.6 xx10^(-5)J`