A body describes simple harmonic motion with amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm (b) 3 cm (c) 0 cm.

When a body describes simple harmonic motion it goes away from its original mean position up to a maximum distance called as the amplitude of the simple harmonic motion, then return towards the mean position and goes to the other extreme position and again then return to the mean position, the total time taken during this complete cycle is called the Time Period

Position of a Particle Undergoing simple harmonic motion with time Period T has been shown in the figure

Here we are given Amplitude of simple harmonic motion as

A = 5 cm = 0.05m

Time Period of simple harmonic motion as

T = 0.2 s

The angular frequency of simple harmonic motion is given as

ω = 2π/T

Where ω is the angular frequency of simple harmonic motion and T is the time period of simple harmonic motion

So here angular frequency is

ω = 2π/0.2s = 10π rad/s

now velocity of a particle undergoing simple harmonic motion is given by the relation

where v is the velocity of particle undergoing simple harmonic motion with amplitude A and angular frequency ω and is at a distance of x from mean position

Acceleration of a particle undergoing simple harmonic motion is given by the relation

a = -ω²x

where a is the acceleration of particle undergoing simple harmonic motion with angular frequency ω and is at a distance of x from mean position

(-ve sign indicates that acceleration is always directed towards the mean position and is in opposite direction to the displacement of particle)

(a) here displacement of particle from mean position is

x = 5 cm = 0.05 m

Here we are given Amplitude of simple harmonic motion as

A = 5 cm = 0.05m

here angular frequency is

ω = 10π rad/s

so putting values in the relation

we get

so velocity of the particle is 0 m/s

to find the acceleration we use the relation

a = -ω²x

so putting the values in the relation we get

a = -(10πs^-1)²×0.05m

= -100π²s^-2×0.05m

= -5π² ms^-2

So acceleration of the particle is 5π² ms^-2 directed towards the mean position

(b) here displacement of particle from mean position is

x = 3 cm = 0.03 m

Here we are given Amplitude of simple harmonic motion as

A = 5 cm = 0.05m

here angular frequency is

ω = 10π rad/s

so putting values in the relation

we get

so velocity of the particle is 0.4π m/s

to find the acceleration we use the relation

a = -ω²x

so putting the values in the relation we get

a = -(10πs^-1)²×0.03m

= -100π²s^-2×0.03m

= -3π² ms^-2

So acceleration of the particle is 3π² ms^-2 opposite to the velocity of the particle directed towards the mean position

(c) here displacement of particle from mean position is

x = 0 cm = 0 m

Here we are given Amplitude of simple harmonic motion as

A = 5 cm = 0.05m

here angular frequency is

ω = 10π rad/s

so putting values in the relation

we get

so velocity of the particle is 0.5π m/s

to find the acceleration we use the relation

a = -ω²x

so putting the values in the relation we get

a = -(10πs^-1)² × 0m

= -100π²s^-2 × 0m

= 0 ms^-2

So acceleration of the particle is 0 ms^-2