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The displacement of a particle varies wi...

The displacement of a particle varies with time according to the relation `y= a sin omega t+b cos omega t`……………

A

the motion is oscillatory but not SHM

B

the motion is SHM with amplitude `a+b`

C

the motion is SHM with amplitude `(a^(2)+ b^(2))`

D

the motion is SHM with amplitude `sqrt((a^2)+(b^2))`

Text Solution

Verified by Experts

The correct Answer is:
D
`y= a sin omega t+b cos omega t`
comparing `y= A sin omega t cos phi +A cos omega t sin phi`
`therefore a= A sin phi " and "b= A cos phi`
`therefore a^(2)+ b^(2)= A^(2) (cos^(2) phi + sin^(2)phi)= A^(2)`
`therefore A= sqrt((a^2)+(b^2))`
`implies y= a sin omega t+ b cos omega t`
`= A sin omega t sin phi + A cos omega t cos phi`
`= A sin (omega t +phi)`
`(dy)/(dx)= A omega cos (omega t +phi)`
and `(d^(2)y)/(dt^(2))= -A omega^(2) sin (omega t+phi)`
`= -omega^(2) (A sin omega t+phi)`
`therefore a = -omega^(2)y`
Hence it is equation of SHM with amplitude `A= sqrt(a^(2)+b^(2))`.
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