Three travelling waves are superimposed. The equations of the wave are
`y_(!)=A_(0) sin (kx-omegat), y_(2)=3 sqrt(2) A_(0) sin (kx-omegat+phi)` and `y_(3)=4 A_(0) cos (kx-omegat)`
find the value of `phi ("given "0 le phi le pi//2)` if the phase difference between the resultant wave and first wave is `pi//4`

Answer

`12.00`

Explanation

`tan((pi)/4)=(BC)/(AC)=(A_(0)(4+3sqrt(2)sin phi))/(A_(0)(1+3sqrt(2)cos phi))`

`impliescos phi-sin phi=1/(sqrt(2))`
Squaring both sides
`impliescos ^(2)phi+sin^(2)phi-2cos phisinphi=1/2`
`implies2sin phi cos phi =1/2`
`impliessin 2 phi=1/2`
`impliesphi=1/2sin^(-1)(1/2)=(pi)/12`