Simple harmonic systemsAQA A-Level Physics: Revision notes

For a mass $m$ on a spring of spring constant $k$ the restoring force is $F = -kx$, so $a = -(k/m)x$ and the motion is simple harmonic with

$T = 2\pi\sqrt{\frac{m}{k}}$

For a simple pendulum of length $l$ (to the centre of the bob) in a field of strength $g$:

$T = 2\pi\sqrt{\frac{l}{g}}$

Note that $T$ for a mass-spring system is independent of $g$ and of amplitude. For a pendulum, $T$ is independent of the mass of the bob. Both equations come from $a = -\omega^2 x$ with $T = 2\pi/\omega$, so $\omega^2 = k/m$ or $g/l$.

Worked example: a 0.250 kg mass on a 40 N m⁻¹ spring has $T = 2\pi\sqrt{0.250/40} = 0.50$ s. Quadrupling the mass doubles $T$ because $T \propto \sqrt{m}$.

For a pendulum the restoring force is the component of weight along the arc, $mg\sin\theta$. This is proportional to displacement only if $\sin\theta \approx \theta$ (with $\theta$ in radians), which is true for small angles (below about 10°). So pendulum motion is only approximately SHM, and only for small amplitudes.

The examiner may describe other oscillators, such as liquid in a U-tube, a floating object or a trolley between springs. The method is always the same: find the restoring force, apply $F = ma$ to get $a = -(\text{constant})\,x$, and read off $\omega^2$ as the constant. All the information needed will be given.

For liquid of total column length $L$ in a U-tube, the restoring force is $2\rho A g x$ and the mass is $\rho A L$, so $\omega^2 = 2g/L$ and $T = 2\pi\sqrt{L/2g}$.

For an undamped oscillator the total energy is constant: $E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}m\omega^2A^2$.

• Potential energy: $E_p = \tfrac{1}{2}kx^2$, zero at equilibrium and maximum at $x = \pm A$.
• Kinetic energy: $E_k = E - E_p = \tfrac{1}{2}m\omega^2(A^2 - x^2)$, maximum at equilibrium and zero at the amplitude.

Against displacement $E_p$ is a parabola opening upwards, $E_k$ a parabola opening downwards, and $E$ a horizontal line. Against time $E_k$ and $E_p$ each vary with period $T/2$, in antiphase, and sum to a constant.

Worked example: $m = 0.400$ kg, $k = 25.0$ N m⁻¹, $A = 0.080$ m. $E = 0.080$ J. At $x = 0.050$ m, $E_p = 0.031$ J, $E_k = 0.049$ J and $v = 0.49$ m s⁻¹.

Real oscillators lose energy because resistive forces (air resistance, friction, viscosity) do work against the motion. This damping transfers energy to the surroundings as thermal energy, so the amplitude falls with each cycle, and the total energy decreases.

• Light damping: amplitude decreases gradually (exponentially) and the period is almost unchanged.
• Heavy damping: amplitude falls faster and the period increases.
• Critical damping: the system returns to equilibrium in the shortest time without oscillating, as in car suspension or door closers.
• Overdamping: returns to equilibrium without oscillating, but slowly.

Mass-spring: hang a mass from a spring on a clamp, displace it a small distance and time the oscillations with a stopwatch, using a fiducial marker (such as a pointer or fixed pin) at the centre of the oscillation. Time at least 10 oscillations, divide to find $T$, and repeat to find a mean. Vary $m$ and plot $T^2$ against $m$: the gradient is $4\pi^2/k$.

Pendulum: use a small dense bob, small angles (below about 10°) and measure the length to the centre of the bob. Vary $l$ and plot $T^2$ against $l$: the gradient is $4\pi^2/g$, so $g$ can be found.

Reduce uncertainty by timing many oscillations and starting the timing as the bob passes the fiducial marker, not at the extremes.