Analyse and interpret the graphical representations of displacement, velocity and acceleration for simple harmonic motion (SHM).
Key Concepts
Simple harmonic motion is a type of periodic motion where the restoring force is directly proportional to the displacement from equilibrium.
The motion can be described by sinusoidal functions of time.
Displacement (\$x\$), velocity (\$v\$) and acceleration (\$a\$) are related through differentiation and integration.
Mathematical Description
The standard equation for displacement in SHM is
\$x(t) = A\cos(\omega t + \phi)\$
where
\$A\$ – amplitude (maximum displacement)
\$\omega\$ – angular frequency (rad s\(^{-1}\))
\$\phi\$ – phase constant
From this, the velocity and acceleration are obtained by differentiation:
\$v(t) = \frac{dx}{dt} = -A\omega\sin(\omega t + \phi)\$
\$a(t) = \frac{dv}{dt} = -A\omega^{2}\cos(\omega t + \phi) = -\omega^{2}x(t)\$
Graphical Relationships
Suggested diagram: Overlaid plots of \$x(t)\$, \$v(t)\$ and \$a(t)\$ for one complete cycle of SHM.
Displacement vs. Time
The displacement graph is a cosine (or sine) wave. Key features:
Maximum and minimum values are \$+A\$ and \$-A\$.
The curve crosses the equilibrium (zero) position twice per cycle.
The period \$T = \frac{2\pi}{\omega}\$ is the time between successive peaks.
Velocity vs. Time
The velocity graph is a sine wave, shifted by a quarter period relative to displacement.
Zero velocity occurs at the extreme positions (\$x = \pm A\$).
Maximum speed \$|v|_{\max}=A\omega\$ occurs when \$x=0\$.
The amplitude of the velocity graph is \$A\omega\$.
Acceleration vs. Time
The acceleration graph is a cosine wave identical in shape to the displacement graph but inverted and scaled.
Acceleration is always directed towards the equilibrium position.
Maximum magnitude \$|a|_{\max}=A\omega^{2}\$ occurs at \$x=\pm A\$.
The acceleration graph has the same period as displacement.
Phase Relationships
Quantity
Phase relative to \$x(t)\$
Amplitude
Displacement \$x(t)\$
Reference (0°)
\$A\$
Velocity \$v(t)\$
\$-90^{\circ}\$ (or \$+270^{\circ}\$)
\$A\omega\$
Acceleration \$a(t)\$
\$180^{\circ}\$ (inverted)
\$A\omega^{2}\$
Interpreting Sample Graphs
Identify the period \$T\$ from any of the three graphs – the distance between two successive peaks.
Determine the amplitude \$A\$ from the displacement graph; verify that the velocity amplitude equals \$A\omega\$ and the acceleration amplitude equals \$A\omega^{2}\$.
Check the phase shift: the velocity curve should cross zero when the displacement curve is at its extreme values.
Confirm that the acceleration curve is a mirror image of the displacement curve (same shape, opposite sign).
Common Misconceptions
Amplitude vs. Maximum Speed: Students often think the maximum speed equals the amplitude. In SHM, \$v_{\max}=A\omega\$, which depends on both amplitude and angular frequency.
Phase Difference: The \$90^{\circ}\$ phase difference between \$x\$ and \$v\$ is sometimes confused with a \$180^{\circ}\$ shift. Remember \$v\$ leads \$x\$ by \$90^{\circ}\$ (or lags by \$-90^{\circ}\$).
Acceleration Direction: Acceleration is always towards the equilibrium, not necessarily opposite to the direction of motion.
Worked Example
A mass‑spring system oscillates with amplitude \$A = 0.05\ \text{m}\$ and angular frequency \$\omega = 10\ \text{rad s}^{-1}\$.
Write the displacement equation (choose \$\phi = 0\$ for simplicity): \$x(t)=0.05\cos(10t)\$.
Determine the period: \$T = \frac{2\pi}{\omega}= \frac{2\pi}{10}=0.628\ \text{s}\$.
State the maximum speed \$v{\max}=A\omega = 0.5\ \text{m s}^{-1}\$ and maximum acceleration \$a{\max}=A\omega^{2}=5\ \text{m s}^{-2}\$.
Summary
SHM is described by sinusoidal functions; displacement, velocity and acceleration are related by differentiation.
All three quantities share the same period; the amplitude of velocity and acceleration scale with \$\omega\$ and \$\omega^{2}\$ respectively.
Phase relationships: \$v\$ is \$90^{\circ}\$ out of phase with \$x\$, and \$a\$ is \$180^{\circ}\$ out of phase with \$x\$.
Graphical analysis provides a quick way to extract \$A\$, \$\omega\$, \$T\$, and to verify the consistency of the three curves.
Practice Questions
Given a displacement graph with \$T = 2\ \text{s}\$ and peak-to-peak amplitude \$0.4\ \text{m}\$, calculate the maximum speed and maximum acceleration.
Sketch the velocity and acceleration graphs for the motion described in Question 1, indicating key points (zero crossings, maxima, minima).
A particle undergoing SHM has \$a(t) = -9\cos(3t)\$. Write the corresponding \$x(t)\$ and \$v(t)\$ expressions, stating the amplitude and angular frequency.