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What is the minimum acceleration for a particle in simple harmonic motion obeying x equals acos2t?
Wiki User
∙ 11y ago
Acceleration is a vector, so it has both magnitude and direction.
A particle in the simple harmonic motion, x = a cos(2t), has an acceleration that varies between positive and negative extremes (+4a and -4a) at the end points of its motion and has zero acceleration as it passes through its center of motion.
Here is the mathematics.
Position ' x ' = a cos(2t)
Velocity ' v ' = dx/dt = -2a sin(2t)
Acceleration = dv/dt = -4a cos(2t)
The minimum instantaneous magnitude of acceleration is zero,
and its most negative instantaneous value is [ -4a ].
Wiki User
∙ 11y ago
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