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Olympiad Oscillations 5.3.1(c)(i)5.3.1(d)5.3.1(e)5.3.2(a)5.3.3(b)(i)
~45 min Difficulty: 8/10
Prior knowledge Basic SHMenergy conservationbinomial expansion.
Diagram description
A vertical glass tube with a fixed magnet at the bottom and a levitating magnet at height h. An axis shows h increasing upwards.
Problem structure
(a), 4, symbolic, Derivation of SHM from a non-linear force. (b), 4, numerical, Application of SHM period to find physical constants. (c), 6, numerical, Energy conservation in a non-linear potential field. (d), 4, numerical, Logarithmic decrement and damping coefficients. (e), 2, qualitative, Resonance in forced oscillations.

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A "magnetic spring" is constructed using two identical small neodymium magnets. One magnet is fixed to the base of a vertical glass tube. The second magnet, of mass $m = 12.0\text{ g}$, is placed inside the tube such that it is repelled by the fixed magnet and levitates at an equilibrium height $h_0$ above it. The repulsive force $F_m$ between the magnets can be modelled by the power law $F_m = \frac{k}{h^n}$, where $h$ is the separation between the centers of the magnets, $k$ is a constant related to the magnetic strength, and $n$ is a dimensionless exponent. (a) At equilibrium, the levitating magnet is at $h_0 = 4.50\text{ cm}$. If the magnet is displaced slightly by a distance $x$ from equilibrium ($x \ll h_0$), show that the resulting motion is approximately simple harmonic and derive an expression for the angular frequency $\omega$ in terms of $n, g,$ and $h_0$. [4] (b) An experimenter measures the period of small vertical oscillations to be $T = 0.350\text{ s}$. Using this data, determine the value of the exponent $n$. [4] (c) The magnet is now pushed down to a height $h_{min} = 3.00\text{ cm}$ and released from rest. Assuming the power law remains valid and ignoring air resistance, calculate the maximum height $h_{max}$ reached by the magnet. [6] (d) In a real setup, the magnet experiences a damping force $F_d = -bv$, where $v$ is the velocity. It is observed that after 10 full oscillations, the amplitude of the motion reduces by 40%. Calculate the damping constant $b$ in $\text{kg s}^{-1}$. [4] (e) If the base of the tube is now vibrated vertically with a very small amplitude $A_{drive}$ at a frequency of $2.86\text{ Hz}$, describe the resulting motion of the levitating magnet and explain why the amplitude of the magnet's oscillation might become unexpectedly large. [2]
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(a) The net force on the levitating magnet is $F_{net} = \frac{k}{h^n} - mg$. At equilibrium ($h=h_0$), $F_{net} = 0$, so $\frac{k}{h_0^n} = mg \implies k = mg h_0^n$. For a small displacement $x$ such that $h = h_0 + x$, the restoring force is: $F_{restoring} = \frac{k}{(h_0+x)^n} - mg = k h_0^{-n} (1 + \frac{x}{h_0})^{-n} - mg$. Using the binomial expansion $(1+\epsilon)^{-n} \approx 1 - n\epsilon$ for $\epsilon \ll 1$: $F_{restoring} \approx mg(1 - n\frac{x}{h_0}) - mg = - \frac{nmg}{h_0}x$. Since $F = ma$, we have $ma = - \frac{nmg}{h_0}x \implies a = - \frac{ng}{h_0}x$. This is the SHM equation $a = -\omega^2 x$, where $\omega = \sqrt{\frac{ng}{h_0}}$. (b) From $T = \frac{2\pi}{\omega}$, we have $T = 2\pi \sqrt{\frac{h_0}{ng}}$. Squaring both sides: $T^2 = \frac{4\pi^2 h_0}{ng}$. Rearranging for $n$: $n = \frac{4\pi^2 h_0}{g T^2}$. Substitute $h_0 = 0.045\text{ m}$, $g = 9.81\text{ m s}^{-2}$, and $T = 0.350\text{ s}$: $n = \frac{4 \times \pi^2 \times 0.045}{9.81 \times 0.350^2} = \frac{1.7765}{1.2017} \approx 1.478$. Rounding to appropriate significant figures, $n \approx 1.48$. (c) We use conservation of energy. The potential energy $U$ consists of gravitational $U_g = mgh$ and magnetic $U_m$. $F_m = -\frac{dU_m}{dh} \implies U_m = -\int \frac{k}{h^n} dh = \frac{k}{(n-1)h^{n-1}}$. Total Energy $E = mgh + \frac{mgh_0^n}{(n-1)h^{n-1}}$. At $h_{min} = 0.030\text{ m}$ and $h_{max}$, the velocity is zero, so $E(h_{min}) = E(h_{max})$. $mg h_{min} + \frac{mg h_0^n}{(n-1)h_{min}^{n-1}} = mg h_{max} + \frac{mg h_0^n}{(n-1)h_{max}^{n-1}}$. Divide by $mg$ and substitute $n=1.478$: $0.030 + \frac{0.045^{1.478}}{0.478 \times 0.030^{0.478}} = h_{max} + \frac{0.045^{1.478}}{0.478 \times h_{max}^{0.478}}$. LHS calculation: $0.030 + \frac{0.01017}{0.478 \times 0.1873} = 0.030 + 0.1136 = 0.1436\text{ m}$. We solve $0.1436 = h_{max} + \frac{0.02128}{h_{max}^{0.478}}$ by iteration or trial. If $h_{max} \approx 0.065\text{ m}$: $0.065 + \frac{0.02128}{0.065^{0.478}} = 0.065 + 0.0788 = 0.1438$ (Close). Refining: $h_{max} \approx 0.0648\text{ m} = 6.48\text{ cm}$. (d) For a damped oscillator, $x(t) = A_0 e^{-\gamma t} \cos(\omega t)$ where $\gamma = \frac{b}{2m}$. After 10 periods, $t = 10T$. The amplitude ratio is $\frac{A_{10}}{A_0} = e^{-\gamma (10T)} = 0.60$. $-10 \gamma T = \ln(0.60) \implies \gamma = \frac{-\ln(0.60)}{10T}$. $\gamma = \frac{0.5108}{10 \times 0.350} = 0.1459\text{ s}^{-1}$. $b = 2m\gamma = 2 \times 0.012 \times 0.1459 = 3.50 \times 10^{-3}\text{ kg s}^{-1}$. (e) The system is a forced oscillator. The natural frequency is $f_0 = 1/T = 1/0.350 \approx 2.86\text{ Hz}$. Since the driving frequency ($2.86\text{ Hz}$) matches the natural frequency of the magnet, resonance occurs. This leads to a maximum transfer of energy from the driver to the magnet, resulting in a large oscillation amplitude. Final Answers: (b) $n = 1.48$ (c) $h_{max} = 6.48\text{ cm}$ (d) $b = 3.50 \times 10^{-3}\text{ kg s}^{-1}$

Marking guidance

(a) [4 marks] - 1 mark: Equating $mg = k/h_0^n$ to find $k$. - 1 mark: Expressing net force for $h_0 + x$ and using binomial expansion correctly. - 1 mark: Identifying the form $a = -Cx$ and stating $C = \omega^2$. - 1 mark: Correct final expression $\omega = \sqrt{ng/h_0}$. (b) [4 marks] - 1 mark: Correct relationship between $T$ and $\omega$. - 1 mark: Rearranging for $n$ algebraically. - 1 mark: Correct substitution of all values in SI units. - 1 mark: Final value $n \approx 1.48$ (accept 1.47-1.49). (c) [6 marks] - 1 mark: Correct integration of force to find magnetic potential energy $U_m \propto h^{1-n}$. - 1 mark: Statement of conservation of energy $E_{tot} = U_g + U_m$. - 1 mark: Correct expression for total energy including the $k$ constant or $mgh_0^n$ term. - 1 mark: Setting $E(h_{min}) = E(h_{max})$. - 1 mark: Numerical calculation of the energy constant (approx 0.144m in head units). - 1 mark: Solving for $h_{max} \approx 6.48\text{ cm}$ (accept 6.4-6.6 cm). (d) [4 marks] - 1 mark: Recalling or deriving the exponential decay of amplitude $A = A_0 e^{-(b/2m)t}$. - 1 mark: Correct time interval $t = 10T = 3.5\text{ s}$. - 1 mark: Correct use of logarithms to solve for $b$. - 1 mark: Final value $3.5 \times 10^{-3}\text{ kg s}^{-1}$ with correct units. (e) [2 marks] - 1 mark: Identifying that the driving frequency equals the natural frequency. - 1 mark: Identifying the phenomenon as resonance.