Physics 9702/43 — October/November 2014

Background Concept

For simple harmonic motion (SHM), the displacement varies sinusoidally with time and repeats after one period $T$. The angular frequency $\omega$ (in $\text{rad s}^{-1}$) is related to the period by

$$\omega = \frac{2\pi}{T}.$$

Here, $2\pi\ \text{rad}$ corresponds to one complete cycle.

Understanding the Question

You are given a displacement–time graph for the vertical oscillations of the magnet. The task is to use the graph to find the period $T$ and hence calculate $\omega$.

Approach

1. Identify two identical points one cycle apart (e.g. two successive maxima).
2. Read the time difference to get $T$.
3. Substitute into $\omega = 2\pi/T$.

Step-by-Step Reasoning

• From the graph description, there is a maximum at $t = 0$ and the next maximum at $t = 0.30\ \text{s}$. Hence

$$T = 0.30\ \text{s}.$$

• Then

$$\omega = \frac{2\pi}{0.30} = 20.94\ \text{rad s}^{-1} \approx 21\ \text{rad s}^{-1}.$$

Key Takeaways

• The period is read from repeated points on the displacement–time graph.
• Angular frequency is found using $\omega = 2\pi/T$.

Common Mistakes

• Using half a cycle (e.g. max to min) as the period.
• Forgetting the factor of $2\pi$ and using $\omega = 1/T$.

Things to Be Careful About

• Choose clearly identifiable points (two maxima or two minima) to reduce reading error.
• Keep units consistent: $T$ in seconds gives $\omega$ in $\text{rad s}^{-1}$.

Background Concept

In SHM, the speed is greatest as the object passes through the equilibrium position (displacement $y = 0$). The maximum speed is related to amplitude $A$ and angular frequency $\omega$ by

$$v_{\max} = \omega A.$$

The maximum kinetic energy occurs at this point:

$$E_{k,\max} = \frac{1}{2}mv_{\max}^2.$$

(Equivalently, the total energy in SHM is $E = \frac{1}{2}m\omega^2A^2$, which equals the maximum kinetic energy.)

Understanding the Question

From the displacement–time graph you read the amplitude. You already found $\omega$ in part (a)(i). With the given mass $m = 0.130\ \text{kg}$, you must show that the maximum kinetic energy is $6.4\ \text{mJ}$.

Approach

1. Read amplitude $A$ from the graph and convert to metres.
2. Use $v_{\max} = \omega A$.
3. Substitute into $E_{k,\max} = \tfrac{1}{2}mv^2$.

Step-by-Step Reasoning

• The graph shows peaks at about $+1.5\ \text{cm}$ and $-1.5\ \text{cm}$, so

$$A = 1.5\ \text{cm} = 0.015\ \text{m}.$$

• With $\omega \approx 21\ \text{rad s}^{-1}$,

$$v_{\max} = \omega A = 21\times 0.015 = 0.315\ \text{m s}^{-1}.$$

• Then

$$E_{k,\max} = \frac{1}{2}(0.130)(0.315)^2 = 6.45\times 10^{-3}\ \text{J} = 6.45\ \text{mJ} \approx 6.4\ \text{mJ}.$$

Key Takeaways

• In SHM, maximum kinetic energy occurs at equilibrium where speed is maximum.
• Use $v_{\max} = \omega A$ and then $\tfrac{1}{2}mv^2$.

Common Mistakes

• Using amplitude in cm without converting to m, giving an answer $10^4$ too large.
• Using the peak-to-peak displacement ($3.0\ \text{cm}$) instead of the amplitude ($1.5\ \text{cm}$).
• Using $E = mv^2$ instead of $E = \tfrac{1}{2}mv^2$.

Things to Be Careful About

• Check units: $\omega$ in $\text{rad s}^{-1}$, $A$ in $\text{m}$ gives $v$ in $\text{m s}^{-1}$.
• Significant figures: $6.4\ \text{mJ}$ is appropriate given graph readings.

Background Concept

Faraday’s law states that an e.m.f. is induced when the magnetic flux linkage changes:

$$\mathcal{E} = -\frac{\mathrm{d}(N\Phi)}{\mathrm{d}t}.$$

The negative sign is Lenz’s law: the induced effect opposes the change that produced it. In a bulk conductor (like an aluminium cup), induced currents circulate in closed loops; these are called eddy currents. Eddy currents cause heating because the conductor has resistance, so electrical energy is dissipated at a rate $P = I^2R$.

Understanding the Question

Replacing cardboard with aluminium introduces a conducting material near the oscillating magnet. The magnet’s motion changes the magnetic flux through the metal, so induction occurs. The question asks you to explain why this causes the amplitude to decrease (i.e. why the motion is damped).

Approach

1. State that magnet motion causes changing flux linkage in the conductor.
2. Use Faraday’s law to conclude an induced e.m.f. and current.
3. Use Lenz’s law to explain the induced magnetic field opposes the magnet’s motion, creating a resistive (damping) force.
4. Link the damping force to energy loss (work done) and heating in the cup.

Step-by-Step Reasoning

• When the magnet moves up and down, the magnetic field pattern relative to the aluminium changes. That means the magnetic flux $\Phi$ through loops in the cup is changing with time.
• By Faraday’s law, a changing flux induces an e.m.f. around those loops. Since aluminium is conducting, this e.m.f. drives circulating eddy currents.
• By Lenz’s law, the magnetic field produced by the eddy currents acts so as to oppose the change in flux. In practice, this opposition manifests as a force opposing the magnet’s velocity (a “magnetic drag”).
  • On the downward motion, the induced field acts to oppose the increase of flux (and so opposes the downward motion).
  • On the upward motion, it opposes the decrease of flux (and again opposes the upward motion).
• Because there is a force opposite to the motion, the magnet does work against this force each cycle. That energy is transferred to the eddy currents and then dissipated as heat in the aluminium (resistive heating).
• As a result, the mechanical energy of the oscillation decreases, so the amplitude steadily falls (damped oscillation).

Key Takeaways

• Changing magnetic flux induces an e.m.f. (Faraday’s law).
• Induced currents create fields that oppose the motion/change (Lenz’s law).
• The opposition does negative work on the oscillator, reducing its energy and amplitude.

Common Mistakes

• Saying “the aluminium attracts the magnet” rather than opposing its motion.
• Describing induction but not linking it to energy dissipation (must mention loss as heat / $I^2R$).
• Confusing the role of the cardboard cup: cardboard is an insulator so eddy currents cannot flow.

Things to Be Careful About

• The damping is strongest when the magnet moves fastest because $\mathrm{d}\Phi/\mathrm{d}t$ is largest then.
• The question states the change in angular frequency is negligible: so you should focus on energy loss (amplitude reduction), not changing $\omega$.

Background Concept

For SHM with angular frequency $\omega$ and amplitude $A$, the total mechanical energy is

$$E = \frac{1}{2}m\omega^2A^2.$$

So if $m$ and $\omega$ are constant (the question says the change in $\omega$ is negligible), then

$$E \propto A^2.$$

This is why damping that reduces amplitude also reduces energy.

Understanding the Question

The magnet starts with the amplitude shown in Fig. 1.2 (about $1.5\ \text{cm}$). With an aluminium cup, after 10 complete oscillations the amplitude is $0.75\ \text{cm}$. You must find how much energy has been lost over those 10 oscillations.

You can use the maximum kinetic energy from part (a)(ii) as the initial total energy $E_0$.

Approach

1. Use $E \propto A^2$ to find the fraction of energy remaining.
2. Multiply that fraction by the initial energy to get final energy.
3. Subtract to get the loss.

Step-by-Step Reasoning

• Initial amplitude $A_0 = 1.5\ \text{cm}$, final amplitude $A_1 = 0.75\ \text{cm}$.
• Energy ratio:

$$\frac{E_1}{E_0} = \left(\frac{A_1}{A_0}\right)^2 = \left(\frac{0.75}{1.5}\right)^2 = (0.5)^2 = 0.25.$$

So the oscillation has $25\%$ of its original energy after 10 oscillations.

• With $E_0 = 6.4\ \text{mJ}$:

$$E_1 = 0.25\times 6.4 = 1.6\ \text{mJ}.$$

• Energy lost:

$$\Delta E = 6.4 - 1.6 = 4.8\ \text{mJ}.$$

Key Takeaways

• In SHM (fixed $m$ and $\omega$), energy scales with $A^2$.
• Halving the amplitude reduces the energy to one quarter.

Common Mistakes

• Assuming energy is proportional to amplitude $A$ rather than $A^2$.
• Using $A_0 = 3.0\ \text{cm}$ (peak-to-peak) instead of amplitude.
• Treating $6.4\ \text{mJ}$ as “energy lost” rather than initial energy.

Things to Be Careful About

• The result depends on $\omega$ being effectively constant (stated in the question).
• Keep consistent units: because you use an energy ratio, cm is acceptable for the amplitude ratio (the conversion cancels).