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`.

Explanation

`(a)` `P` is moving up i.e. has positive velocity, particle velocity,
`v_(p)=-v((dy)/(dx))`……`(1)` `v=`wave velocity
`(dy)/(dx)=` slope at point `P=tan60^(@)=sqrt(3)`
As `v_(P)` and `(dy)/(dx)` are both positive hence `v` must be negative, hence the wave is moving in negative `x`-axis.
`(b)` The equation of the wave
`y=asin(omegat-kx+delta)`......`(2)`
`omega=2pin` and `k=(omega)/(v)=(2pi)/(lambda)=(pi)/(2)cm^(-1)`
`a=4xx10^(-3)m=0.4cm` At `t=0`, `x=0` and `y=2xx10^(-3)m=0.2cm`
Substituting in `(2)`
`2sqrt(2)xx10^(-3)=4xx10^(-3)sin delta`, which gives `delta=(pi)/(4)` or `(3pi)/(4)`
Also at `t=0`, `x=1.5cm`. And `y=0`
`0=asin(0-(3pi)/(4)+delta)` which gives `delta=(3pi)/(4)`
hence, `delta=(3pi)/(4)`
`20sqrt(3)=-vxxtan60^(@)` ,
`v=-20cm//s`(-sign because wave is moving in negative direction)
Hence, `n=v//lambda=5Hz`.
Hence, `omega=2pixx5=10pi`
Substituting the values of `a`, `omega`, `k` and `delta`
We get,
`y=0.4 sin (10pi t-(pi)/(2)x+(pi)/(6))`
or `y=0.4sin2pi(5t-(x)/(4)+(1)/(12))`.