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## SHM as a Projection of Circular 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:

• aN​=2sin2(ωt+ϕ)
• aL​=2cos2(ωt+ϕ)

aN and aL 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.

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## Kinetic Energy of a Particle in SHM

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

1/22A2cos2(ωt+ϕ)

1/22(A2x2)

Therefore, the Kinetic Energy = 1/2​2A2cos2(ωt+ϕ)=1/2​2(A2x2)

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## Time Period of SHM

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.

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## Quantitative Analysis of SHM

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

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.

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## Angular 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
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## Solutions of Differential Equations of SHM

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.

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## Simple Harmonic Motion Equation and its Solution

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)