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`.

Options

A. `y 5xx10^(-3) sin(10pit-50pix-(pi)/(8))`
B. `y 4xx10^(-3) sin(10pit-50pix-(pi)/(8))`
C. `y (5xx10^(-3)(sin(10pit(50pix-(pi)/(8))`
D. `y (4xx10^(-3)(sin(10pit(50pix-(pi)/(8))`

Answer

d

Explanation

(a). At `P:` slope of tangent
`=(dy)/(dt)=tan60^(@)=sqrt(3)`
particle velocity
`v_(p)=-v(dy)/(dt)implies20sqrt(3)=-vsqrt(3)`
`implies |v|=20 cm//s=(1)(5)m//s`
Hence, the wave is travelling in negative `x`-direction with velocity `20 cm//s`.
(b). Frome graph, amplitude `A=4xx10^(-3) m`
wave length `lambda=(5.5-1.5)=4xx10^(-2)m`
wave number
`K=(2pi)/(lambda)=(2pi)/(4xx10^(-2))=50pi m^(-1)`
Angular frequency `omega=kv`
`=50pixx(1)/(5)=10pi`
Hence, equation can be written as
`y=A sin(omegat+kx+phi)`
`=(4xx10^(-3))sin(10pit+50pix+phi)`
at `=0,x=0`
`2sqrt2xx10^(-3)=4xx10^(-3)sin(phi)`
`impliessinphi(1)/sqrt(2), phi=(pi)/(40, (3pi)/(4)`
particle is moving up at `t=0,x=0`
Hence, `phi=(pi)/(4)`
Hence, equation is
`y=(4xx10^(-3))sin(10pit+50pix+(pi)/(4))`