The equation of a travelling plane sound wave has the form `y=60 cos (1800t-5.3x)`, where y is in micrometers, t in seconds and x in meters. Find
(a). The ratio of the displacement amplitude with which the particle of the medium oscillate to the wavelength,
(b).the velocity oscillation amplitude of particles of the medium and its ratio to the wave propagation velocity , (c ).the oscillation amplitude of relative deformation of the medium and its relation to the velocity oscillation amplitude of particles of the medium, (d). the particle acceleration amplitude.

Explanation

Comparing the given equation `y=60 cos (1800t-5.3x)` with the wave equation `y omegaacos omegat-kx0,`
`omega=1800 rad//s, a=60xx10^(-6)m,k=5.3` per metre.
As `k=(2pi)/(lambda), lambda=(2pi)/(k)=(2pi)/(5.3)m`
(a). Displacement amplitude of particcle of the medium=amplitude of the wave `=60xx10^(-6) m`.
Required `"ratio"=("Displacement amplitude")/("wavelength")`
`=(60xx10^(-6)m)/(2pi/5.3)m`
`=(60)/(2pi)xx10^(-6)xx5.3=5.06xx10^(-5)`
(b). Velocity oscillation amplitude`=` Magnitude of the velocity of a particle of medium when passes through its equilibrium position `aomega`
velocity of propagation of the wave `=f lambda=(omega)/(2pi)xxlambda=(omega)/((2pi/lambda))=(omega)/(k)`
Required ratio
`(aomega)/((omega/k))=ka=5.3xx60xx10^(-6)=3.18xx10^(-4)`
(c) velocity of particle `dy//dt aomega cos omegat-kx`
Acceleration of the particle
`(d^(2)y)/(dt^(2)=aomega[-sin(omegat-kx)]xxomega=-aomega^)2) sin(omegat-kx)`
Maximum value of `|sin omegat-kx | 1`
Thus, maximum acceleration of the particle `=`particle accelration amplitude `aomega^(2)=60xx10^(-6)(1800)^(2)`
`=194.4 m//s^(2)SSS`