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

(i) Negative x
(ii) `y = 0.4 sin (10 pi t + (pi)/(2) x + (pi)/(4))`
(x,y are in cm)
(iii) `1.6 xx 10^(-5) J`

Explanation

(i) For the particle P
`(dely)/(delt)=-v((dely)/(delx))rArr +20sqrt(3)=v(sqrt(3))`
`rArrv=-20cm//s`
(along negative x-axis)
(ii) Equation of wave
`y=Asin(omegat+kx+phi)`
at`t =0,x=0,y=2sqrt(2),A=4`
`rArr4=2sqrt(2)sinphirArrphi=(pi)/(4),lamda=5.5-1.5=4cm`
`f=(v)/(lamda)=(20 cm//s)/(4cm)=5Hz`
`:.y=4sin(10pit+(pix)/(2)+(pi)/(4))`
(iii) Energy carried in one wavelength
`E=(1)/(2)muA^(2)omega^(2)lamda`
`=(1)/(2)xx(50)/(1000)xx(4xx10^(-2))^(2)xx(10pi)^(2)xx(4)/(100)`
`=16 pi^(2) xx 10^(-5) J`