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Three travelling waves are superimposed....

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`

A

`pi/6`

B

`pi/3`

C

`pi/12`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C
Given, `y_(1)=A_(0)sin(kx- omegat )`
`" "y_(2)=3sqrt(2)A _(0)sin(kx-omega t+phi),`
and `y_(3) =4A_(0)cos(kx- omegat)`,

These waves can be represented by phase diagram as shown.
From phase diagram,
`tan((pi)/(4) )=(BC)/(AC)=(A_(0)(4+3sqrt(2)sinphi))/(A_(0)(1+3sqrt(2)cosphi))`
`rArr" "4+3 sqrt(2)sinphi=1+3sqrt(2)cosphi`
`rArr" "cosphi-sinphi=(1)/(sqrt(2))`
Squaring both sides, we get
`cos^(2)phi+sin^(2 )phi-2sinphicosphi =(1)/(2)`
`1-2sinphicos phi=(1)/(2)" or "2phi=(pi)/(6)rArrphi=(pi)/(12)`
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