Q: A body executes SHM, such that its velocity at the mean position is 1 ms^{-1} and acceleration at extreme position is 1.57 ms^{-2}. Calculate the amplitude and the time period of oscillation.

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Sol: $\large \frac{a_{max}}{v_{max}} = \frac{\omega^2 A}{\omega A} $

$\large \frac{1.57}{1} = \omega $

ω = 1.57 rad/sec

$\large T = \frac{2\pi}{\omega} = \frac{2 \times 3.14}{1.57} = 4 sec $

Aω = 1 i.e., A(1.57) = 1

or A = 1/1.57

Amplitude A = 0.637 m.