A plane wave propagates along positive x-direction in a homogeneous medium of density `p=200 kg//m^(3)`. Due to propagation of the wave medium particle oscillate. Space density of their oscillation energy is `E=0.16pi^(2) J//m^(3)` and maximum shear strain produced in the mendium is `phi_(0)=8pixx10^(-5)`. if at an instant, phase difference between two particles located at points `(1m,1m,1m)` and `(2m, 2m, 2m,)` is `Deltatheta=144^(@)`, assuming at `t=0` phase of particle at `x=0` to be zero,
wave length is

Options

A. `2.5 m`
B. `5 m`
C. `10 m`
D. `6 m`

Answer

a

Explanation

Since, the wave is a plane travelling wave, intensity at every point will be the same,
since, initial phase of particle at `x=0` is zero and the wave is travelling along positive `x-` deirection equation of the wave will be of the form
`delta=a sin omega (t-(x)/(v))` ...`(i)`
Let intensity of the wave be `I`, then space density of oscillation energy of medium particles will be equal to
`E=(I)/(v)`
But, `I=2pi^(2)n^(2)a^(2)p=0.16pi^(2)J//m^(3)`
`a^(2)n^(2)=4xx10^(-4)`
or, `an=0.02`
shear strain of the medium is
`phi=(d)/(dx)delta`
Differentiating Eq. (i),
`phi=-(aomega)/(v)cosomega(t-(x)/(v))`
Modulus of shear strain `f` will be maximum when
`cosomega(t-(x)/(v))=+-1`
`:.`maximum shear srain `8pixx10^(-5)`
`phi_(0)=(aomega)/(v)`
but it is equal to
`(aomega)/(v)=8pixx10^(-5)`
where
`omega=2pin`
`an=4vxx10^(-5)` ....`(iii)`
Solving Eqs. (ii) and (iii), `v=500 m//s`
since, the wave is travelling along positive `x-`direction, there, phase difference is give by
`Deltatheta=2pi(Deltax)/(lambda)`
`Deltax=(x_(2)-x_(1))=(2-1)m=1m`
`lambda=(2piDeltax)/(Deltatheta)=2.5 m`
But `v=nlambda`,therefore,
`n=(v)/(lambda)=200 Hz`
substituting `n=200 Hz` in Eq. (ii),
`a=1xx10^(-4) m`
Angular frequency,`omega=2pi n=400 pi rad//s`. substituting all these values in Eq. (i),
`Delta=10^(-4) sinpi(400t-0.8x)m`
since, due to propagation of the wave, shear strain is produced in the medium, the wave is a plane transverse wave.