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A particle executes simple harmonic moti...

A particle executes simple harmonic motion with an angular velocity and maximum acceleration of `3.5 rad // s ` and `7.5 m // s^(2)` respectively. Amplitude of the oscillation is .......... .

A

0.36

B

0.28

C

0.61

D

0.53

Text Solution

Verified by Experts

The correct Answer is:
C
`x=A sin omegat`
`:. a=(d^(2)x)/(dt^(2))=-Aomega^(2) sin omegat`
`:."Maximum acceleration" |a_(max)|=Aomega^(2)`
Now `Aomega^(2)=7.5`
`A=(7.5)/(omega^(2))=(7.5)/((3.5)^(2))=0.61`
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