`x=x_(1)+x_(2)` (where `x_(1)=4 cos omega t ` and `x_(2)=3 sin omega t`) is the equation of motion of a particle along x-axis. The phase different between `x_(1)` and x is
A
`37^(@)`
B
`53^(@)`
C
`90^(@)`
D
none of these
Text Solution
Verified by Experts
The correct Answer is:
A
`x=x_(0) sin (omegat+phi)=x_(0)sin omegat cos phi+x_(0) cos omegatsin phi`. Comparing with given equation thus `x_(0)cos phi=3` and `x_(0)sin phi=4` Dividing we get `tan phi=4/3` or `phi=53^(@)` `x_(1)=4 cos omegat =4 sin (omegat+90^(@))` `Delta theta=90^(@)-53^(@)=37^(@)`
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