SHM Phasors

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SHM Phasors

This post is only an extension to the previous article on AC Phasors. The reader is advised to read the article on AC Phasors here first, to gain an elementary idea on phasors. In this article, we consider phasors used in describing Simple Harmonic Motion.

Simple Harmonic Motion:

The motion governed by a force proportional to the displacement of the particle from the mean position and directed towards the mean position is called simple harmonic motion. i.e. for an ideal particle of mass `m`, we have,

` F = -kx`

i.e.   ` m \frac{\text{d}^{2}x}{\text{dt}^{2}} = -kx `

Suppose we have the mass particle `m` attached to a spring that applies force according to the above equation. We stretch it from the mean position to some distance say `a` and then release it. 




The differential equation of this motion is then,

` \frac{\text{d}^{2}x}{\text{dt}^{2}} = - \frac{k}{m}x`

Since both `k` and `m` are positive quantities, we can let `\frac{k}{m} = w^{2}`, for some `w` (this `w` later acts as the angular frequency). The above equation then transforms into,

`\frac{\text{d}^{2}x}{\text{dt}^{2}} = - w^{2}x`

`\frac{\text{d}^{2}x}{\text{dt}^{2}} + w^{2}x = 0`

This is a linear differential equation of the second order. The complete solution of the equation is of the form,

` x = c_{1}cos(wt) + c_{2}sin(wt)`

For velocity, we can differentiate the above expression. Thus,

` v = -c_{1}w sin(wt) + c_{2}w cos(wt)`

If we apply the initial conditions of `x = a` &  `v = 0`, at `t = 0`, we get `c_{1} = a` and `c_{2} = 0`. Thus,

` x = a cos (wt) `

Here `a` is called the amplitude. Since`-1 \leq cos(wt) \leq 1`, the amplitude is the maximum displacement from the mean position. 

Now, as we had discussed in the article on AC Phasors, we can write the above as follows:

` x = Re(A_{0}e^{jwt}) `

where `A_{0} = a angle 0 ` is the phasor of the sinusoid `a cos(wt)`. Thus, we have successfully related the displacement to a rotating phasor in a complex domain. This can be seen in the below animation.







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