Simple Harmonic Oscillations – Cambridge A‑Level Physics 9702
What is Simple Harmonic Motion (SHM)?
Imagine a playground swing. When you push it and let it go, it swings back and forth with a regular rhythm. That regular rhythm is simple harmonic motion – a motion where the restoring force is proportional to the displacement and always points back to the equilibrium position.
Key Equations
| Quantity | Formula |
|---|
| Displacement | \$x(t)=A\cos(\omega t+\phi)\$ |
| Velocity | \$v(t)=-A\omega\sin(\omega t+\phi)\$ |
| Acceleration | \$a(t)=-A\omega^2\cos(\omega t+\phi)\$ |
| Angular frequency | \$\omega=2\pi f=\sqrt{k/m}\$ |
Graphical Representations
When you plot these quantities against time, you get beautiful curves that tell you everything about the motion.
- Displacement vs. Time – a cosine wave that starts at its maximum, crosses zero, goes to a minimum, and repeats. The amplitude \$A\$ is the peak height.
- Velocity vs. Time – a sine wave that is 90° out of phase with displacement. The maximum speed is \$A\omega\$.
- Acceleration vs. Time – a cosine wave but inverted (negative). It is proportional to displacement: \$a = -\omega^2 x\$.
📈 Key Observations:
- All three graphs have the same period \$T=2\pi/\omega\$.
- The velocity graph is shifted by \$\pi/2\$ (90°) ahead of displacement.
- The acceleration graph is shifted by \$\pi\$ (180°) and has a larger amplitude by a factor \$\omega^2\$.
- When the displacement is zero, the velocity is at its maximum (and vice versa).
- When the displacement is at a maximum or minimum, the acceleration is at its maximum (in magnitude) and velocity is zero.
Analogy: The Pendulum & the Spring
Think of a spring attached to a mass. Pull it down, release it, and watch it oscillate. The spring’s force is \$F=-kx\$, exactly the restoring force needed for SHM. Similarly, a pendulum (for small angles) behaves like SHM because the restoring torque is proportional to the angle.
Example Problem (Graph Interpretation)
You are given a displacement‑time graph of a mass on a spring. The graph shows a peak at \$x=0.05\text{ m}\$ and a period of \$0.8\text{ s}\$.
- Amplitude \$A = 0.05\text{ m}\$.
- Period \$T = 0.8\text{ s}\$ → angular frequency \$\omega = 2\pi/T = 7.85\text{ rad s}^{-1}\$.
- Maximum velocity \$v_{\max}=A\omega = 0.05\times7.85 = 0.393\text{ m s}^{-1}\$.
- Maximum acceleration \$a_{\max}=A\omega^2 = 0.05\times7.85^2 = 3.08\text{ m s}^{-2}\$.
Exam Tips – Information Box
📌 Remember:
- Use the correct phase shift when comparing graphs.
- Check the units – displacement in metres, time in seconds, velocity in m s⁻¹, acceleration in m s⁻².
- When asked for “maximum speed” or “maximum acceleration”, look for the peak values on the respective graphs.
- In questions about energy, remember that kinetic energy is highest when velocity is highest (mid‑point of motion).
- For graphs with missing axes labels, assume standard units unless stated otherwise.
Quick Quiz – Test Your Understanding
- If the displacement graph has a period of 2 s, what is the angular frequency?
- At what times does the velocity reach zero in one full cycle?
- Which graph shows the greatest amplitude: displacement or acceleration? Why?
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Good luck, and remember: the rhythm of SHM is all around – from clocks to musical instruments. Keep exploring the patterns and you’ll master the graphs in no time! 🎯