The equation of a plane standing wave in a homogeneous elastic medium has the form `xi= a cos kx. Cos omegat`. Plot `:`
(a) `xi` and `delta xi // delta x ` as functions of `x` at the moments `t=0` and `t=T//2`, where `T` is the oscillation period,
`(b)` the distribution of density `rho(x)` of the medium at the moments `t=0` and `t=T//2` in the case of longitudinal oscillation ,
`(c)` the velocity distribution of particles of the medium at the moment `t=T//4` , indicate the directions of velocities at the antinodes, both for longitudinal and transverse oscillation.

Explanation

We are given
K `xi = a cos k x omegat `
so ` ( deltaxi)/( deltax) = -a k sin k x cos omegat `and ` ( delta xi )/( deltat)=-a omega cos k x sin omega t `
Thus `( xi) _(t=0)= a cos k x, ( xi)_(t=T//2)= -a cos kx `
`((deltaxi)/( deltax))_(t=0)=a k sin kx, ((delta xi )/( deltax))_(t=T//2)=-a k sin kx `
`(a)` The graphs of `(xi)` and `((delta xi)/( deltax))` are as shown in figure of the book `(p.332)`.
`(b)` We can calculate the density as follows `:`
Take a parallelopiped of cross section unity and length `dx` with its edges at `x` and `x+ dx`.
After the oscillation the edge at `x` goes to `x+ xi(x)` and the edge at `x+ dx` goes to `x+dx+xi(x+dx)`
`=x+ dx+ xi(x)+(deltaxi)/( deltax) dx` . Thus the volume of the element `(` originally `dx)` becomes
`(1+(deltaxi)/( deltax))dx`
and hence the density becomes `rho=(rho_(0))/(1+(delta xi)/( deltax))`.
On substituting we get for the density `rho(x)` the curves shown in figure, referred to above.
`(c)` The velocity `v(x)` at time `t=T//4` is
`((delta xi)/(deltat))_(t=T//4)=-a omega cos kx `
On plotting we get the figure.