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The diplacement of a particle varies wit...

The diplacement of a particle varies with time according to the relation `y=a"sin"omegat+b " cos"omegat`.

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
(d) Accoding to the question , the displacement
`y=a "sin"omegat+b "cos"omegat`
Let `a=A"sin"phi " and "b=A"cos" phi`
Now , `a^(2)+b^(2)=A^(2)"sin"^(2)phi+A^(2)"cos"^(2)phi`
`=A^(2)impliesA=sqrt(a^(2)+b^(2))`
`therefore y=A"sin"phi, "sin" omegat+A"cos"phi."cos"omegat`
`=A"sin"(omegat+phi)`
`implies(dy)/(dt)=Aomega"cos"(omegat+phi)`
`implies(d^(2)y)/(dt^(2))=-Aomega^(2)"sin"(omegat+phi)=-Ayomega^(2)=(-Aomega^(2))y`
`implies(d^(2)y)/(dt^(2))prop(-y)`
Hence, it is an equation of SHM with amplitude, `A=sqrt(a^(2)+b^(2))`.
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