Category: 1. Simple Harmonic Motion

The particle is at position P at t = 0 and revolves with a constant angular velocity (ω) along a circle. The projection of P on the diameter along the x-axis (M). At the later time (t) the particle is at Q. Now its projection on the diameter along the x-axis is N.

As the particle P revolves around in a circle anti-clockwise its projection M follows it up moving back and forth along the diameter such that the displacement of the point of projection at any time (t) is the x-component of the radius vector (A).

x = A cos ( ωt + Φ) . . . . . . . (1)

y = A sin ( ωt + Φ) . . . . . (2)

Thus, we see that the uniform circular motion is the combination of two mutually perpendicular linear harmonic oscillation.

It implies that P is under uniform circular motion, (M and N) and (K and L) are performing simple harmonic motion about O with the same angular speed ω as that of P.

P is under uniform circular motion, which will have centripetal acceleration along A (radius vector)\overrightarrow{{{a}_{c}}}=A{{\omega }^{2}}ac​​=Aω2 (towards the centre)

It can be resolved into two components:

• a_N​=Aω²sin²(ωt+ϕ)
• a_L​=Aω^₂cos²(ωt+ϕ)

a_N and a_L acceleration corresponding to the points N and L respectively.

In the above discussion, the foot of projection on the x-axis is called horizontal phasor.

Similarly, the foot of the perpendicular on the y-axis is called vertical phasor. Already we know the vertical and horizontal phasor will execute the simple harmonic motion of amplitude A and angular frequency ω. The phases of the two SHM differ by π/2.

Kinetic Energy = 1/2​mv² [Since,v²=A²ω²cos²(ωt+ϕ)]

1/2mω2A2cos2(ωt+ϕ)

1/2mω²(A²−x²)

Therefore, the Kinetic Energy = 1/2​mω²A²cos²(ωt+ϕ)=1/2​mω²(A²−x²)

The coefficient of t is ω.

So the time period T = 2π/ω

ω =2π/T = 2πf

ωt = angular frequency of SHM.

From the expression of particle position as a function of time:

We can find particles, displacement ( \overrightarrow{x} ),(x),velocity ( \overrightarrow{v})(v) and acceleration as follows.

Let us consider a particle executing Simple Harmonic Motion between A and A₁ about passing through the mean position (or) equilibrium position (O). Its analysis is as follows

SHM about Position O

Displacement	x = -A	x = 0	x = +A
Acceleration	|a| = Max	a = 0	|a| = max
Speed	|v| = 0	|v| = Max	|v| = 0
Kinetic energy	KE = 0	KE = Max	KE = 0
Potential energy	PE = Max	PE = Min	PE = Max

Equation of Position of a Particle as a Function of Time

Let us consider a particle, which is executing SHM at time t = 0, the particle is at a distance from the equilibrium position.

Necessary conditions for Simple Harmonic Motion

A body free to rotate about an axis can make angular oscillations. For example, a photo frame or a calendar suspended from a nail on the wall. If it is slightly pushed from its mean position and released, it makes angular oscillations.

Conditions for an Angular Oscillation to be Angular SHM

The body must experience a net Torque that is restoring in nature. If the angle of oscillation is small, this restoring torque will be directly proportional to the angular displacement.

Τ ∝ – θ

Τ = – kθ

Τ = Iα

α = – kθ

This is the differential equation of an angular Simple Harmonic Motion. Solution of this equation is angular position of the particle with respect to time.

θ=θ0​sin(ω0​t+ϕ)

Then angular velocity,

ω=θ0​.ω0​cos(ω0​t+ϕ)

θ₀ – amplitude of the angular SHM

Example:

• Simple pendulum
• Seconds pendulum
• The physical pendulum
• Torsional pendulum

The differential equation for the Simple harmonic motion has the following solutions:

• x=Asinωt (This solution when the particle is in its mean position point (O) in figure (a)

• x0​=Asinϕ (When the particle is at the position & (not at mean position) in figure (b)

• x=Asin(ωt+ϕ) (When the particle at Q at in figure (b) (any time t).

These solutions can be verified by substituting this x values in the above differential equation for the linear simple harmonic motion.

Consider a particle of mass (m) executing Simple Harmonic Motion along a path x o x; the mean position at O. Let the speed of the particle be v₀ when it is at position p (at a distance no from O)

At t = 0 the particle at P (moving towards the right)

At t = t the particle is at Q (at a distance x from O)

With a velocity (v)