The following equation gives the displacement u at time t for a particle at a distance x:
`y=0.01 sin 500pi(t-x//30)`
where all are in SI units.
Find (i) the wavelength, (ii) the speed of the wave, (iii) the velocity amplitude of the particle of the medium and (iv) the accelration amplitude of the particle of the medium.

Explanation

Comparing the equation with the standard from of the wave equation
`y=a sin(omegat-kx), where omega =2pif and k = 2pi//lambda`
Given that (i)`omega=500pi or f=250 Hz`
and `(2pi)/(lambda)=(500p0i)/(30)` or `lambda=(3)/(25)m=0.12 m`
(ii). `:. C=flambda` or C=250xx(3)/(25)=30 m//s`
(iii). `v("particle velocity") = (dy)/(dt)=aomega cos(omegat-kx)`
`:. v_(max)=aomega=0.01xx500pi=15.7 m//s`
(iv). f(particle acceleration)
`=(dv)/(dt)=aomega^(2) sin(omegat-kx)`
`:. f_(max)=aomega^(2)=0.01xx(500pi)^(2) = 2.47xx10^(4) m//s^(2)`