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The instantaneous displacement x of a pa...

The instantaneous displacement `x` of a particle executing simple harmonic motion is given by `x=a_1sinomegat+a_2cos(omegat+(pi)/(6))`. The amplitude `A` of oscillation is given by

A

`sqrt(a_1^2+a_2^2+2a_1a_2cos((pi)/(6)))`

B

`sqrt(a_1^2+a_2^2+2a_1a_2cos((pi)/(3)))`

C

`sqrt(a_1^2+a_2^2-2a_1a_2cos((pi)/(6)))`

D

`sqrt(a_1^2+a_2^2-2a_1a_2cos((pi)/(3)))`

Text Solution

Verified by Experts

The correct Answer is:
D


`x=a_1sinomegat+a_2cos(omegat+(pi)/(6))`
`=a_1sinomegat+a_2sin(omegat+(pi)/(6)+(pi)/(2))`
`=a_1sinomegat+a_2sin(omegat+(2pi)/(3))`
`A^2=a_1^2+a_2^2+2a_1a_2cos((2pi)/(3))`
`a_1^2+a_2^2-2a_1a_2cos((pi)/(3))`
`A=sqrt(a_1^2+a_2^2-2a_1a_2cos((pi)/(3)))`
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