Simple acceleration in Harmonic motion !

Classical Mechanics Level 1

The acceleration of a particle performing simple harmonic motion is 12 cm per sec at a distance of 3 cm from the mean positions. Its time period is:-


The answer is 3.14.

Parag Zode by Parag Zode , 24, India

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1 solution

A K
Apr 30, 2014

a = − ω 2 x a = - \omega^{2} x

a = − ( 2 π T ) 2 x a = -(\frac{2\pi}{T})^{2}x

a x = − 4 π 2 T 2 \frac{a}{x} = \frac{-4\pi^{2}}{ T^{2}}

− 12 3 = − 4 π 2 T 2 \frac{-12}{3} = \frac{-4\pi^{2}}{T^{2}}

Note that the acceleration in simple harmonic motion is always regarded as being negative.

T 2 = π 2 T^{2} = \pi^{2}

Therefore, in seconds:

T = π \boxed{T = \pi}

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