Velocity and acceleration of particle performing simple harmonic motion

We know that when a body is set up in vibrations from its original/mean position by external force it come backs to its mean position due to action of restring force present in the body and said to be performing the linear simple harmonic motion. Whereas the linear motion of object along the circumference of circle with uniform velocity is known as uniform circular motion. The acceleration of particle in linear SHM is given as,

Where ω is angular velocity and x is displacement of particle,given as

ω = 2π/T

(T is period of SHM)

Let’s obtain the differential equation for linear simple harmonic motion……..!

Consider a particle of mass m performing linear SHM with angular velocity ω. Let x be the displacement of particle from its mean position.

Then the restoring force acting on the particle is given as,

F= -Kx……..(1)

Also by Newton’s 2nd law of motion,

F = ma ……..(2)

From equations (1) and (2) we get,

ma = -kx……..(3)

Now the acceleration of the particle is given as,

a = dv/dt

but velocity, v= dx/dt

then acceleration can be written as,

Now we shall derive the formulae for acceleration and velocity in linear SHM…….!

1) Acceleration in SHM:-

We know the differential equation of SHM is,

From above equation we can conclude that acceleration in SHM is directly proportional to the displacement, and acts in opposite direction of displacement.

2) Velocity in SHM:-

Using equation for acceleration, we can find out the velocity of particle in SHM as follows,

Then equation (2) becomes,

This is an equation for velocity of particle performing linear SHM, where is angular frequency.

Some important values for acceleration and velocity:-

1.) Acceleration:-

When particle is at extreme position, x=A

∴ accelerationmax = – ω2 A

When particle is at mean position, x=0

∴ accelerationmin = 0

2) Velocity:-

When particle is at mean position, x=0

∴Vmax = Aω

When particle is at extreme position, x = 0

∴Vmin = 0

Let’s solve some numerical to understand in more details…..!

Ex:1) A particle performs SHM with period of 3π s. If amplitude of SHM is 12 cm, find the velocity and acceleration of particle at distance of 6 cm from mean position.

Solution:

Here, T = 3π s, A = 12 cm, x = 6 cm

Now we have,

ω = 2π/T

∴ω = 2π/3π

∴ω = 2/3 rad/s

1.) Acceleration is given as,

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