In simple harmonic motion,the displacement x = A cos $\omega$t and the acceleration a = -$\omega$²x.If A = 0.25 m and the period is 0.32 s,the acceleration when t = 0.12 s is A)zero B)+3.9 m/s² C)-3.9 m/s² D)+6.8 m/s² E)-6.8 m/s²

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The Answer of In simple harmonic motion,the displacement x = A cos $\omega$t and the acceleration a = -$\omega$²x.If A = 0.25 m and the period is 0.32 s,the acceleration when t = 0.12 s is A)zero B)+3.9 m/s² C)-3.9 m/s² D)+6.8 m/s² E)-6.8 m/s²

In Simple Harmonic Motion,the Displacement X = a Cos ω\omegaω T and the Acceleration a = -

In Simple Harmonic Motion,the Displacement X = a Cos $\omega$ T and the Acceleration a = -