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

Options

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. Thee motions is `SHM` with amplitude `sqrt(a^(2)+b^(2))`

Answer

D

Explanation

According to the question, the displacement
`y = a sin omegat +b cos omega t`
Given `x = a sin omegat + b cos omega t …(i)`
Let `a = A cos phi …(ii)`
and `b = A sin phi …(iii)`
Squaring and adding (ii) and (iii), we get
`a^(2) +b^(2) =A^(2) cos^(2) phi +A^(2) sin^(2) phi =A^(2)`
`=A^(2) rArr A = sqrt(a^(2)+b^(2))`
`y = A sin phi. sin omegat +A cos phi. cos omegat`
`= A sin (omegat +phi)`
`(dy)/(dt) = A omega cos (omegat +phi)`
`(d^(2)y)/(dt^(2)) =- A omega^(2) sin (omegat +phi)`
`=- A y omega^(2) = (-A omega^(2))y`
`rArr (d^(2)y)/(dt^(2)) alpha -(-y)`
Hence, it is an equation of `SHM` with amplitude
`A = sqrt(a^(2)+b^(2))`