Chapter 1 Simple Harmonic Motion

Simple harmonic motion (SHM) is a simple and common type of oscillatory motion. It is a model which is widely used in modelling systems due to its simplicity.

In general, an object will move under SHM where its acceleration is:

proportional to its displacement, but
in the opposite direction.

The force causing this acceleration is often termed a restoring force as it acts to push the object back to its starting point.

1.1 A simple example of SHM

Consider a block on a spring (Figure 1.1)

$A mass on a spring, stretched distance $x$ past its equilibrium length $x_0$$

Figure 1.1: A mass on a spring, stretched distance $x$ past its equilibrium length $x_0$

By Hooke’s law, the spring exerts a force on the block proportional to its displacement $x$, but in the opposite direction, pushing the block back to its equilbrium position, shown mathematically in Equation (1.1):

\[\begin{equation} F_x = -kx \tag{1.1} \end{equation}\]

In this example, $F_x$ is considered a restoring force, while $k$ is the force constant of the spring.

Applying Newton’s Second Law to this problem, we can obtain the mathematical description of the system:

\[\begin{equation} \begin{aligned} F_x = ma_x \\ -kx = m\frac{\textrm{d}^2 x}{\textrm{d} t^2} \end{aligned} \end{equation}\]

… and through rearrangement and combination with Equation (1.1) we obtain the description of how this mass will move (Equation (1.2)):

\[\begin{equation} \frac{\textrm{d}^2 x}{\textrm{d} t^2} = -\frac{k}{m}x \tag{1.2} \end{equation}\]

The general form of this expresssion for any system can be considered as shown in Equation (1.3)

\[\begin{equation} \frac{\textrm{d}^2 x}{\textrm{d} t^2} = -Cx \quad\mathrm{or}\quad\ddot{x} = -Cx \tag{1.3} \end{equation}\]

…where $C$ is a positive constant which depends on the system and represents a ratio of the elastic ($k$) and inertial ($m$) contributions within the system.

1.1.1 Key terms

Period: The time $T$ for one complete oscilation back and forth (units s)
Frequency: The reciprocal of the period; $f = \frac{1}{T}$, units s^-1 or Hz.

1.2 Positioning in SHM

SHM can be described by a general equation of motion, defining the position ($x$) of the oscillating mass using a cosine function (Equation (1.4))

\[\begin{equation} x = A \cos (\omega t + \delta) \tag{1.4} \end{equation}\]

The parameters in this equation are:

$A$: The amplitude of the oscillation
$\omega t + \delta$: Phase of motion
$\delta$: Phase constant

For any single oscillator, the time origin can always be chosen so that $\delta = 0$. For two or more oscillators there will generally be a phase difference between them (i.e. they will not always be at the same ‘zero’ position at time zero - Figure 1.2)

Figure of position vs time for two oscilators

Figure 1.2: An illustration of the changing position of two oscillators with respect to time, with a relative phase shift of 2 radians.

1.3 Velocity in SHM

To find the velocity of the oscillating mass, we can simply find the first derivative of its position with respect to time (Equation (1.5)):

\[\begin{equation} \begin{array}{rcl} v & = & \frac{\textrm{d}x}{\textrm{d}t}\\ &=& -A \omega \sin (\omega t + \delta)\\ \end{array} \tag{1.5} \end{equation}\]

A quick inspection of this shows that the velocity $v$ is maximised when $x$ is at a minimum; i.e. as the object passes through its equilibirum position.

1.4 Acceleration in SHM

Again, to find the acceleration, we find the second derivative of its position with respect to time (Equation (1.6)):

\[\begin{equation} \begin{array}{rcl} a & = & \frac{\textrm{d}v}{\textrm{d}t} = \frac{\textrm{d}^2 x}{\textrm{d}t^2}\\ &=& -A \omega^2 \cos (\omega t + \delta)\\ \end{array} \tag{1.6} \end{equation}\]

…or, to use the Newtonian “dot” notation (Equation (1.7)):

\[\begin{equation} a = \ddot{x} = -\omega^2 x \tag{1.7} \end{equation}\]

If we now compare this with Equation (1.2) we can see that we have an expression for $\omega$ for the oscillating mass $m$ on a spring of force constant $k$ (Equation (1.8)):

\[\begin{equation} \begin{array}{rcl} \omega^2 &=& \frac{k}{m}\\ \omega &=& \sqrt{\frac{k}{m}} \end{array} \tag{1.8} \end{equation}\]

1.5 Comparing displacement, velocity and acceleration

When we now compare the displacement, velocity and acceleration we make a number of observations. Firstly, they are all sinusoidal functions; variously sine and cosine functions. However, when we overlay these we have a better indication of how they interrelate (Figure 1.3)

Comparing the changes of position, velocity and acceleration with time for a harmonic oscillator. Note that when $x$ is at zero, $v$ is maximised, while $a$ is at a maximum when $v$ is zero. The relative amplitudes of each of the waves is given.

Figure 1.3: Comparing the changes of position, velocity and acceleration with time for a harmonic oscillator. Note that when $x$ is at zero, $v$ is maximised, while $a$ is at a maximum when $v$ is zero. The relative amplitudes of each of the waves is given.

We can now make a few more observations:

When the displacement $x$ is at a maximum ($x_\textrm{max}$), the velocity $v$ is zero while the acceleration is at its maximum but negative with respect to displacement ($a= -a_\textrm{max}$)
When the displacement $x$ is zero, the velocity $v$ is at its maximum value ($v = \pm v_\textrm{max}$) and the acceleration is zero.
The pattern repeats with each period; namely $x_0$ (displacement at time $t = 0$) is equal to the displacement $x_T$ (displacement after one period of oscillation, $T$), and the same for the acceleration and velocity.
In general, $x_t = x_{t+T}$; the displacement at time $t$ is equal to the displacement at the time $t$ plus one period of oscillation, $T$.

This allows us to compare the displacement, velocity and acceleration at four points in the oscillation (Table 1.1):

Table 1.1: Relating the displacement, velocity and acceleration at different times in the oscillation for a simple harmonic oscillator.
Time	Displacement, $x$	Velocity, $v$	Acceleration, $a$
$t = 0$	$x_0 = A$	$v_0 = 0$	$a_0 = -a_{\textrm{max}}$
$t = \frac{T}{4}$	$x_{\frac{T}{4}} = 0$	$v_{\frac{T}{4}} = -v_\textrm{max}$	$a_{\frac{T}{4}} = 0$
$t = \frac{T}{2}$	$x_{\frac{T}{2}} = -A$	$v_{\frac{T}{2}} = 0$	$a_{\frac{T}{2}} = a_{\textrm{max}}$
$t = \frac{3T}{4}$	$x_{\frac{3T}{4}} = 0$	$v_{\frac{3T}{4}} = v_\textrm{max}$	$a_{\frac{3T}{4}} = 0$
$t = T$	$x_T = x_0 = A$	$v_T = v_0 = 0$	$a_T = a_0 = -a_{\textrm{max}}$

1.6 Initial conditions

We mentioned in Section 1.5 that the displacement, velocity and acceleration expressions were based on sinusoial functions, and each function had a scaling factor $A$ (the amplitude of the oscillation) and a phase component $\delta$. In most problems, we wish to determine the value of these constants. In order to determine these, we establish the initial conditions of the oscillation.

In Figure 1.3 we defined our displacement at $+A$ which set up the rest of the problem. However, we will not always be so fortunate. For a general case, we then need to be more discerning.

We can establish expressions for both the amplitude and the phase component by setting $t = 0$ in our general expressions (Equation (1.9)):

\[\begin{equation} \begin{array}{rclcrcl} x &=& A \cos (\omega t + \delta) & \rightarrow & x_0 &=& A \cos (\delta)\\ v &=& -A \omega \sin (\omega t + \delta)& \rightarrow & v_0 &=& -A \omega \sin ( \delta) \end{array} \tag{1.9} \end{equation}\]

We now treat these as simultaneous equations to find $\delta$ and $A$ (Equation (1.10))¹:

\[\begin{equation} \begin{array}{rcl} \tan \delta = \dfrac{\sin \delta}{\cos \delta} = -\dfrac{v_0}{\omega x_0} & \textrm{and} & A^2 = x_0^2 + \dfrac{v_0^2}{\omega^2} \end{array} \tag{1.10} \end{equation}\]

1.7 Frequency and angular frequency

In Section 1.5 we stated that the nature of the oscillation meant that it repeats after every oscillation; mathematically $x(t) = x(t + T)$; the position $x$ at time $t$ is equal to the position at time $(t+T)$.

When we apply this to the velocity, we obtain the following expression:

\[\begin{equation} \begin{array}{rcl} v(t) &=& v(t+T) \\ A \cos (\omega t + \delta) &=& A \cos (\omega (t+T) + \delta) \\ &=& A \cos ([\omega t + \delta] + \omega T) \end{array} \end{equation}\]

Due to the cyclic nature of a cosine function, $\cos (\alpha) = \cos (\alpha + 2\pi)$, this must therefore mean (Equation (1.11)):

\[\begin{equation} \omega T = 2\pi \hspace{10pt} \textrm{or} \hspace{10pt} \omega = \frac{2\pi}{T} \tag{1.11} \end{equation}\]

This gives us a way to think about $\omega$; its connection to circular motion (the clue is the $2\pi$!). It can be thought of as the angular frequency, with units radians s^-1, and an oscillation of $2\pi$ radians corresponds to one period of oscillation.

Additionally, since the frequency of the oscillation $f$ is the reciprocal of the period of oscillation ($f = \frac{1}{T}$), the angular frequency can be rewritten as $\omega = 2\pi f$, and $f = \frac{\omega}{2\pi}$.

For the spring system we discussed in Section 1.1, we stated that the angular frequency $\omega = \sqrt{\frac{k}{m}}$. Therefore we can obtain an expression for the frequency of our oscillator (Equation (1.12))

\[\begin{equation} f = \frac{1}{T} = \frac{1}{2\pi}\sqrt{\frac{k}{m}} \tag{1.12} \end{equation}\]

Inspection of this equation reveals the behaviour of our oscillator:

If we have a stiffer spring (larger $k$), we expect the frequency $f$ to increase,
If we use an oscillator with larger mass (larger $m$), we would expect the frequency ($f$) to decrease.
The frequency (and therefore period) of simple harmonic oscillation is independent of amplitude.²

1.8 SHM and circular motion

We mentioned an “angular frequency” for SHM; this would appear to suggest behaviour akin to circular motion. It is therefore worth exploring our descriptions of circular motion.

Imagine a point mass moving in a circle (Figure 1.4). For convenience, we imagine this using Cartesian $x-y$ axes, shown in Figure 1.4.

A particle moving in a circle of radius $A$ can be assumed to have an instantaneous linear velocity $v$. The $x$ and $y$ components of the motion are found from trigonometry of the radius $A$ and the angle $θ$.

Figure 1.4: A particle moving in a circle of radius $A$ can be assumed to have an instantaneous linear velocity $v$. The $x$ and $y$ components of the motion are found from trigonometry of the radius $A$ and the angle $θ$.

The particle of mass $m$ is moving in a circle of radius $A$ with instantaneous linear velocity $v$; the radius makes an angle $\theta$ with the $x$-axis. We now look at how its position maps onto each of the axes:

The angular velocity of the particle is $\omega$; found via $\frac{v}{A}$
- We can then describe $\theta$ in terms of $\omega$:
- $\theta = \omega t + \delta$ ($\delta$ is the angle at time $t=0$)
The particle’s position on the $x$-axis is therefore found via:
- $x = A \cos \theta = A \cos (\omega t + \delta)$
- This corresponds with the expression for SHM for a particle moving in a linear fashion (Equation (1.4)).

We can also consider how its position maps onto the $y$-axis:

The position on the $y$-axis is found via:
- $y = A \sin \theta = A \sin (\omega t + \delta) \equiv A \cos(\omega t + [\delta - \frac{\pi}{2}])$
- This once again corresponds with the expression for SHM for a particle moving in a linear fashion.
- The $y$-component of the motion is $\frac{\pi}{2}$ out of phase wiht the $x$-component

This illustrates that circular motion is a combination of two perpendicular SHM oscillations of the same frequency and amplitude, but a relative phase of $\frac{\pi}{2}$.

1.9 Energy in SHM

As with all isolated systems, the total energy $E$ of the simple harmonic oscillator is constant, however the contributions from potential energy ($U$) and KE vary with time.

\[\begin{equation} E = KE + U = \textrm{constant} \end{equation}\]

Let’s go back to the condition for SHM; there is a restoring force proportional to the displacement:

\[\begin{equation} F = -kx \end{equation}\]

Knowing that the potential energy is the first derivative of the force, we can integrate this force expression (with respect to $x$) to get back to the energy statement:³

\[\begin{equation} \begin{array}{rcl} U &=& \int F \mathrm{d}x \\ & =& \frac{1}{2}kx^2 \hspace{10pt} [+C] \end{array} \tag{1.13} \end{equation}\]

However, we already have an expression for how the displacement, $x$, varies with time (Equation (1.4)); let’s now substitute this into the result from Equation (1.13):

\[\begin{equation} U = \frac{1}{2}kA^2 \cos^2 (\omega t + \delta) \tag{1.14} \end{equation}\]

We can also generate an expression for the kinetic energy; remember that kinetic energy can be found from $\frac{1}{2}mv^2$; so we use the expression for $v$ given in Equation (1.5):

\[\begin{equation} \begin{array}{rcl} KE &=& \frac{1}{2} mv^2 \\ & =& \frac{1}{2} m A^2 \omega^2 \sin^2 (\omega t + \delta) \end{array} \tag{1.15} \end{equation}\]

We can simplify this using Equation (1.6) for a particle on a spring, where $\omega^2 = \frac{k}{m}$:

\[\begin{equation} KE = \frac{1}{2}kA^2 \sin^2 (\omega t + \delta) \tag{1.16} \end{equation}\]

Combining the result of Equations (1.14) and (1.16) we find the result in Equation (1.17):

\[\begin{equation} \begin{array}{rcl} E_\textrm{total} &=& U + KE \\ & =& \frac{1}{2}kA^2 \cos^2 (\omega t + \delta) + \frac{1}{2}kA^2 \sin^2 (\omega t + \delta)\\ & =& \frac{1}{2}kA^2 \left[ \cos^2 (\omega t + \delta) + \sin^2 (\omega t + \delta)\right]\\ &=& \frac{1}{2}kA^2 \end{array} \tag{1.17} \end{equation}\]

This result tells us that the total energy in a simple harmonic oscillation is proportional to the square of the amplitude.

Comparing how the kinetic energy (KE) and potential energy (U) of an harmonic oscillator varies with displacement $x$ about the equilibrium position.

Figure 1.5: Comparing how the kinetic energy (KE) and potential energy (U) of an harmonic oscillator varies with displacement $x$ about the equilibrium position.

1.9.1 Points to bear in mind

$U = U_\mathrm{max}$ at $x = ±x_\mathrm{max}$
$KE = KE_\mathrm{max}$ at $x = 0$
$U_\mathrm{average} = KE_\mathrm{average} = \frac{1}{2}E_\mathrm{total}$

Note that we use the trigonometric identity $\cos^2 \alpha + \sin^2 \alpha = 1$ to find $A$↩︎
A caveat to this is for large amplitudes where other factors start to affect the behaviour. But this is then no longer simple harmonic motion!↩︎
The constant of integration will evaluate to zero from the starting condition $U = 0$ at zero displacement.↩︎

Time	Displacement, \(x\)	Velocity, \(v\)	Acceleration, \(a\)
\(t = 0\)	\(x_0 = A\)	\(v_0 = 0\)	\(a_0 = -a_{\textrm{max}}\)
\(t = \frac{T}{4}\)	\(x_{\frac{T}{4}} = 0\)	\(v_{\frac{T}{4}} = -v_\textrm{max}\)	\(a_{\frac{T}{4}} = 0\)
\(t = \frac{T}{2}\)	\(x_{\frac{T}{2}} = -A\)	\(v_{\frac{T}{2}} = 0\)	\(a_{\frac{T}{2}} = a_{\textrm{max}}\)
\(t = \frac{3T}{4}\)	\(x_{\frac{3T}{4}} = 0\)	\(v_{\frac{3T}{4}} = v_\textrm{max}\)	\(a_{\frac{3T}{4}} = 0\)
\(t = T\)	\(x_T = x_0 = A\)	\(v_T = v_0 = 0\)	\(a_T = a_0 = -a_{\textrm{max}}\)