6.1 Simple Harmonic Motion

Oscillations are periodic variations of a quantity about some equilibrium. This could be an object moving back and forth, a voltage going up and down, or any other physical variable. Let’s look at a system we have seen before, a mass on a horizontal spring. We are going to study the motion of the mass in much more detail than earlier. The only force acting on the mass is the spring restoring force, which is given by Hooke’s law:

\boldsymbol{F}=-kx\hat{\boldsymbol{\imath}}.

(6.1)

By Newton II, we can write

a_{x}=\frac{\operatorname{\mathrm{d}}\!^{2}x}{\operatorname{\mathrm{d}}\!t^{2}% }=-\frac{k}{m}x.

(6.2)

The general solution to this differential equation is

x(t)=A_{1}\cos(\omega t)+A_{2}\sin(\omega t),

(6.3)

where

\omega=\sqrt{\frac{k}{m}}.

(6.4)

Systems where displacement is a sinusoidal function of time are said to exhibit simple harmonic motion. This motion is characteristic of any system where the force is oppositely proportional to displacement. We will see more examples of this later.

We can simplify the equation for $x(t)$ using a trigonometric addition formula. If we let $A_{1}=x_{m}\cos\phi$ , $A_{2}=x_{m}\sin\phi$ , where $\phi$ is some angle. Then we have

	$\displaystyle x(t)$	$\displaystyle=x_{m}\cos\phi\cos(\omega t)+x_{m}\sin\phi\sin(\omega t)$		(6.5)
		$\displaystyle=x_{m}\cos(\omega t+\phi).$		(6.6)

We still have two constants $x_{m}$ and $\phi$ rather than $A_{1}$ and $A_{2}$ to specify the particular solution. Since cosine has a maximum value of 1, $x_{m}$ must represent the maximum amplitude of the oscillation. $\phi$ is the initial phase of the oscillation, i.e. the point on the cosine curve where the mass is at $t=0$ . Calculating the derivatives of $x(t)$ , we find

	$\displaystyle v(t)$	$\displaystyle=\frac{\operatorname{\mathrm{d}}\!x}{\operatorname{\mathrm{d}}\!t% }=-x_{m}\omega\sin(\omega t+\phi)$		(6.7)
	$\displaystyle a(t)$	$\displaystyle=\frac{\operatorname{\mathrm{d}}\!^{2}x}{\operatorname{\mathrm{d}% }\!t^{2}}=-x_{m}\omega^{2}\cos(\omega t+\phi)=-\omega^{2}x(t).$		(6.8)

Thankfully we have recovered the equation of motion (6.2) for the acceleration. The two constants can be found using two initial conditions, usually $x(0)$ and $v(0)$ . For example, let’s suppose $x(0)=0$ and $v(0)=v_{0}$ , then we have

	$\displaystyle x(0)=x_{m}\cos\phi=0\implies\phi=\pm\frac{\pi}{2}$		(6.9)
	$\displaystyle v(0)=-x_{m}\omega\sin\phi=v_{0}\implies\phi<0,$		(6.10)

so $x_{m}=\frac{v_{0}}{\omega}$ , $\phi=\frac{\pi}{2}$ .

6.1.1 Period and Frequency of Oscillation

$\omega$ has units of $\mathrm{rad}\text{\,}{\mathrm{s}}^{-1}$ (the same as angular velocity!), and we call it the natural or resonant (angular) frequency. Let’s work out the period of oscillation. It should be equal to the time taken for the phase to change by $2\pi$ .

$\displaystyle\omega t+\phi$	$\displaystyle=\omega(t+T)+\phi$	(6.11)
$\displaystyle\implies\omega T$	$\displaystyle=2\pi$	(6.12)
$\displaystyle T$	$\displaystyle=\frac{2\pi}{\omega}.$	(6.13)

Linear frequency $f$ is a more familiar unit, being measured in cycles per second or $\mathrm{Hz}$ . It is related to the period by $f=1/T$ , and therefore to angular frequency by

\omega=2\pi f.

(6.14)

6.1.2 Motion of a Pendulum

Let’s look at a pendulum on a string of length $L$ now. We are assuming that the string is massless and ignoring the effects of air resistance. Given that the arc length of the pendulum is related to the angular displacement by $s=L\theta$ and that the restoring force is $F=-mg\sin(\theta)$ , we have the equation of motion:

F=m\frac{\operatorname{\mathrm{d}}\!^{2}s}{\operatorname{\mathrm{d}}\!t^{2}}=-% mg\sin(\theta).

(6.15)

This is a nonlinear differential equation. We will simplify this by assuming that the angle $\theta$ is small so $\sin(\theta)\approx\theta$ (small-angle approximation). Then the equation of motion becomes

\frac{\operatorname{\mathrm{d}}\!^{2}s}{\operatorname{\mathrm{d}}\!t^{2}}=-g% \theta=-\frac{g}{L}s.

(6.16)

Note that this has the same form as equation 6.2, so the trajectory will have the same form! What we have effectively done is assume that the displacement $s$ and the restoring force $F$ are pointing in a straight line, not along the arc of a circle, so that the force becomes identical to Hooke’s law.

s(t)=s_{m}\cos(\omega t+\phi),

(6.17)

where $\omega$ in this case is given by

\omega=\sqrt{\frac{g}{L}}.

(6.18)

Example 6.1.

As an exercise, let’s rederive this result by looking at the torques on the pendulum rather than the forces.

The pendulum, being a point mass, has a moment of inertia $I=mL^{2}$ . Then the torque on the pendulum is

\displaystyle\tau=rF\sin\theta=-mgL\sin\theta\approx-mgL\theta.

(6.19)

Using $\tau=I\alpha$ , we get

$\displaystyle\alpha=\frac{\operatorname{\mathrm{d}}\!^{2}\theta}{\operatorname% {\mathrm{d}}\!t^{2}}$	$\displaystyle=-\frac{mgL}{I}\theta$	(6.20)
$\displaystyle\frac{\operatorname{\mathrm{d}}\!^{2}\theta}{\operatorname{% \mathrm{d}}\!t^{2}}$	$\displaystyle=-\frac{mgL}{mL^{2}}\theta$	(6.21)
$\displaystyle\frac{\operatorname{\mathrm{d}}\!^{2}\theta}{\operatorname{% \mathrm{d}}\!t^{2}}$	$\displaystyle=-\frac{g}{L}\theta.$	(6.22)

Solving this, we get

\displaystyle\theta(t)=\theta_{m}\cos(\omega t+\phi),

(6.23)

which is the same solution as before where $s_{m}=L\theta_{m}$ and $\omega=\sqrt{\frac{g}{L}}$ . Note that we have to be quite careful here, as the angular frequency $\omega$ is not the angular velocity $\dot{\theta}$ , since the latter changes with time.

What happens if we don’t use the small-angle approximation? Unfortunately, equation 6.15 cannot be solved analytically. Fortunately, it is easy to solve numerically! Analysing the solution is outside the scope of these notes, but to put it simply, as the maximum amplitude of the oscillation increase, the resonant frequency decreases. I.e. $\omega$ becomes a function of $s_{m}$ (or $\theta_{m}$ ).

6.1.3 SHM in Multiple Dimensions

There is another example of simple harmonic motion that we have seen already. Think back to uniform circular motion. An object moving with UCM has a trajectory of the form

\theta(t)=\omega t+\phi,

(6.24)

where $\omega$ is the angular velocity and $\phi$ is the angular displacement at $t=0$ . If we describe this motion in cartesian coordinates, the trajectory takes the form

\boldsymbol{r}(t)=r\cos(\omega t+\phi)\hat{\boldsymbol{\imath}}+r\sin(\omega t% +\phi)\hat{\boldsymbol{\jmath}},

(6.25)

so we see that UCM is really SHM in two dimensions! If we look at the object dead-on from the $y$ or $x$ direction, we would see identical motion to that of a mass on a spring. For UCM, the motion in the two perpendicular directions are perfectly out of phase (out of phase by one quarter turn, or $\pi/2$ ).

Example 6.2.

Consider a pendulum of mass $m$ and length $L$ which is swinging around in uniform circular motion. What is the frequency of the oscillation?

The radial acceleration is $a_{r}=\dot{\theta}^{2}r=\omega^{2}L\sin\theta$ . Therefore, we have

	$\displaystyle mg\sin\theta=m\omega^{2}L\sin\theta$		(6.26)
	$\displaystyle\implies\omega=\sqrt{\frac{g}{L}},$		(6.27)

which is the same frequency as a normal pendulum.