shows the position of a medium particle at t=0, supporting a simple harmonic wave travelling either along or opposite to the positive x-axis.

(a) write down the equation of the curve. (b) find the angle `theta` made by the tangent at point P with the x-axis. (c ) If the particle at P has a velocity `v_(p) m//s`, in the negative y-direction, as shown in figure, then determine the speed and direction of the wave. (d) find the frequency of the wave. (e) find the displacement equation of the particle at the origin as a function of time. (f) find the displacement equation of the wave.

Explanation

(a). From given curve: amplitude of the wave `A=10 cm` and wavelength `lambda=12 cm`
`y=A sin(kx+-omegat+phi)=A sin((2pi)/(lambda)x+-(2pi)/(T)t+phi)`….(i)
At `t=0, y=A sin(kx+phi)=10 sin(pi)/(6)x+phi)`
At `x=0, y=-5 cm=10 sin phi`
`sin phi=-(1)(2) impliesphi=-(pi)/(6)`
hence `y=0.10(m)sin(pi)/(6)x-(pi)/(6))`
(b). `tantheta =(dy)/(dx)=0.10xx(pi)(6) cos(pi)/(6)x-(pi)/(6))`
At `x=6 m`
`(dy)/(dx)tantheta=(pi)/(600cos(pi-(pi)/(6))=(pi)/(60)(-cos(pi)/(6))`
`=-(pi)/(60).sqrt(3)/(2)=(-pisqrt(3))/(120)`
`theta=tan^(-1)(-(pisqrt(3))/(120))`
(c ). nParticle velocity
`v_(p)=v(dy/dx)`
`-v_(p)=-v(-pisqrt(3))/(120))`
`v=(-120)/sqrt(3pi)v_(p)=(-40sqrt(3))/(pi)v_(p)` Along negative `x-`axis.
(d). wave velocity
`v=f lambda`
`impliesf=(v)/(lambda0=(40sqrt(3)v_(p))/(pixx12)`
`=(10sqrt(3))/(3pi)v_(p)s^(-1)`
(e). hence displacement equation of medium particle at origin at `x=0`, from eq... (ii)
`y=10(cm)sin[(20sqrt(3))/(3)v_(p)t-(pi)/(6)]`
(f). As wave is travelling towards negative x-directin hence positive sign should be used between kx and `omegat` in eq. ..(i)
The equation of the wave,
`y=0.1(m)sin[(pi)/(6)x(m)+omegat-(pi)/(6)]`
here `omega=2pi f=2pi(10sqrt(3))/(3pi)v_(p)=(20sqrt(3))/(3)v_(p)`
Hence displacement equation of the wave
`y=0.1(m)sin[(pi)/(6)x(m)+(20sqrt(3))/(3)v-(p)t-(pi)/(6)]`...(ii)