# A-level Mathematics/OCR/M3/SHM
## SHM Equations of Motion
### Defining Equation of Motion (in terms of acceleration and displacement)

By definition, a particle performs simple harmonic motion (SHM) when
* the magnitude of its acceleration ${\ddot {x}}$ is proportional to the magnitude of its displacement $x$ from some fixed point, and
* its acceleration and displacement are in opposite directions.
More formally, SHM occurs whenever the equation of motion ${\ddot {x}}=-kx$ holds, where $k$ is a positive constant. By convention, we write $k=\omega ^{2}$ (we will explain the significance of $\omega$ later) so the equation becomes ${\ddot {x}}=-\omega ^{2}x$.
Simple harmonic oscillators are easy to find in the real world. The undamped oscillation of a mass attached to an elastic string or spring is just one example in the realm of mechanics. Before studying such examples, let us first examine the characteristics of SHM using the abstract model described below.
Consider the following oscillator with an oscillation amplitude of $a$. Let $x$ denote its displacement from the origin, and ${\ddot {x}}$ its acceleration.
Let us restate the defining equation for SHM:
Recall that the algebraic variables representing displacement and its derivatives wrt time represent one-dimensional vector quantities (i.e. they all lie on a line). If the value of the variable is positive, then its vector points in the positive direction; if it is negative, its vector points in the other direction.
Observe that when the $x=0$, then ${\ddot {x}}=0$. By Newton's Second Law, the resultant force $F$ acting on the oscillator is zero at this instance.
When $x>0$ (i.e. to the right of the origin), the resultant force points to the left, towards the origin. When $x<0$, the resultant force points to the right, towards the origin. Because the resultant force always points towards the origin whenever the oscillator is not at the origin, the resultant force can be termed a restoring force. The behaviour of the restoring force suggests that the speed of the oscillator is maximum at the origin and is zero at the two ends where it changes direction (i.e. $v=0$ when $x=\pm a$).

### Equation of Motion, in terms of velocity, amplitude and displacement

Now, let us rewrite the defining SHM equation in terms of the velocity $v$ and displacement $x$. We make the following substitution: ${\ddot {x}}={\frac {dv}{dt}}={\frac {dv}{dx}}{\frac {dx}{dt}}=v{\frac {dv}{dx}}$. It follows that
| | ${\ddot {x}}$ | $=-\omega ^{2}x\,$ |
| $\Rightarrow$ | $v{\frac {dv}{dx}}$ | $=-\omega ^{2}x\,$ |
| $\Rightarrow$ | $\int v\,dv$ | $=-\omega ^{2}\int x\,dx$ (separating variables) |
Using $v=0$ and $x=\pm a$ as one pair of limits for the integral, we get
| | $\int _{0}^{v}v\,dv$ | $=-\omega ^{2}\int _{\pm a}^{x}x\,dx$ |
| $\Rightarrow$ | $\left[{\frac {1}{2}}v^{2}\right]_{0}^{v}$ | $=-\omega ^{2}\left[{\frac {1}{2}}x^{2}\right]_{\pm a}^{x}$ |
| $\Rightarrow$ | $v^{2}=\omega ^{2}(a^{2}-x^{2})\,$ |
Now, it is clear that the speed is maximum when $x=0$ (in which case $v=\pm \omega a$), and $v=0$ when $x=\pm a$.

### Equation of Motion, in terms of displacement, amplitude and time

The significance of the constant $\omega$ will become clear once we express the defining SHM equation in terms of the displacement $x$ and time $t$:
| | ${\ddot {x}}$ | $=-\omega ^{2}x\,$ |
| $\Rightarrow$ | ${\frac {d^{2}x}{dt^{2}}}+\omega ^{2}x$ | $=0\,$ |
This is just a second-order ordinary differential equation with constant coefficients. The characteristic equation (as a function of $m$) is given by:
$m^{2}+\omega ^{2}=0\Rightarrow m=\pm {\sqrt {-\omega ^{2}}}\Rightarrow m=\pm \mathrm {i} \omega$
Thus, the general solution to the differential equation may be written as:
$x=A_{1}\cos(\omega t)+A_{2}\sin(\omega t)\,$, or equivalently, $x=A_{3}\cos(\omega t+\varepsilon )\,$.
To solve for the pairs of constants $\{A_{1},A_{2}\}$ or $\{A_{3},\varepsilon \}$, we need a pair of initial conditions describing the SHM.
If the particle starts from rest at the maximum displacement $x=a$, then we substitute $x=a$ and ${\frac {dx}{dt}}=0$ at $t=0$ to get $A_{1}=a$, $A_{2}=0$ and $A_{3}=a$, $\varepsilon =0$. The solution becomes $x=a\cos(\omega t)\,$.
On the other hand, if the oscillation begins at minimium displacement, then we obtain $x=a\sin(\omega t)=a\cos \left(\omega t-{\frac {\pi }{2}}\right)\,$.
In general, we have
where the constant $\varepsilon$ depends on the starting position of the oscillator.
These equations show that $\omega$ is just the constant angular velocity (or angular frequency) associated with the SHM. The following figure illustrates this relationship between SHM and uniform circular motion (with an angular velocity of $\omega$). The displacement $x$ of the simple harmonic oscillator is really just the $x$-coordinate of the particle performing uniform circular motion. We can therefore think of SHM as a projection of uniform circular motion onto the diameter of the circle.
The simple harmonic oscillator completes one oscillation whenever it traverses twice the end-to-end distance (i.e. a distance of $4a$). This corresponds to one revolution around the circle in the case of uniform circular motion. Thus, the period of oscillation in SHM is given by

## Example: Mass on an Elastic String

We wish to determine if a mass suspended on a light (i.e. massless) elastic string will demonstrate SHM.
Consider a light elastic string with natural length $l$ and modulus of elasticity $\lambda$ suspended vertically from a fixed point O. The other end of the string is attached to a particle of mass $m$.
Let the point E denote the equilibrium position of the particle. At E, the particle is at rest (i.e. the resultant force acting on it is zero). Let $e$ denote the extension of the string when the particle is at E. Considering the vertical forces on the particle gives:
${\frac {\lambda }{l}}e=mg\Rightarrow e={\frac {mgl}{\lambda }}$.
The displacement $x$ of the mass is always measured from the equilibrium position E, since this is the point where ${\ddot {x}}=0$ and hence, the resultant force $F=0$.
Suppose the mass experiences a small displacement $x$. Applying Newton's Second Law produces
| | $mg-T$ | $=m{\ddot {x}}$ |
| $\Rightarrow$ | $mg-{\frac {\lambda }{l}}(e+x)$ | $=m{\ddot {x}}$ |
| $\Rightarrow$ | $mg-{\frac {\lambda }{l}}e-{\frac {\lambda }{l}}x$ | $=m{\ddot {x}}$ |
| $\Rightarrow$ | $-{\frac {\lambda }{l}}x$ | $=m{\ddot {x}}$ (since $mg={\frac {\lambda }{l}}e$) |
| $\Rightarrow$ | ${\ddot {x}}$ | $=-{\frac {\lambda }{ml}}x$ |
which is the defining equation of motion for SHM, with $\omega ^{2}={\frac {\lambda }{ml}}$. Note that the particle performs SHM only when the string remains taut, i.e. $a<e$.

## Summary

You are expected to know the following formulas, as they will not be provided in the examination:
| Formula | Description |
| ${\ddot {x}}=-\omega ^{2}x$ | SHM Defining Equation of Motion (in terms of acceleration and displacement) |
| $v^{2}=\omega ^{2}\left(a^{2}-x^{2}\right)$ | SHM Equation of Motion, in terms of velocity, amplitude and displacement |
| $x=a\cos(\omega t+\varepsilon )\$ | SHM Equation of Motion, in terms of displacement, amplitude and time |
| $T={\frac {2\pi }{\omega }}$ | SHM Period of Oscillation |