# 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. - Physics

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.

#### 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  –5π2 m/s2 and velocity is 0.

Concept: Force Law for Simple Harmonic Motion
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#### APPEARS IN

NCERT Class 11 Physics
Chapter 14 Oscillations
Exercises | Q 24.1 | Page 361