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Numerical

A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is 5 cm.

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#### Solution

Amplitude, A = 5 cm = 0.05 m

Time period, T = 0.2 s

For displacement, x = 5 cm = 0.05 m

Acceleration is given by:

`"a" = -omega^2x`

`= -((2pi)/"T")^2x`

`= - ((2pi)/0.2)^2xx0.05`

`= -5 pi^2 "m/s"^2`

Velocity is given by:

`"V" = omegasqrt("A"^2 - x^2)`

`= (2pi)/"T" sqrt((0.05)^2 - (0.05)^2)`

= 0

When the displacement of the body is 5 cm, its acceleration is –5π^{2} m/s^{2} and velocity is 0.

Concept: Force Law for Simple Harmonic Motion

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