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

A particle executes simple harmonic motion with an angular velocity an

1.	A particle executes simple harmonic motion with an angular velocity and maximum acceleration of 3.5 rad // sand 7.5 m // s^(2) respectively. Amplitude of the oscillation is .......... .
Answer» 0.36 0.28 0.61 0.53 Solution :`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`