Simple harmonic systems (A-level only)

AQA·Physics·3.6.1.3

Mass–spring oscillations

A mass attached to a spring can move back and forth about its equilibrium position, where the net force on it is zero. When the mass is displaced from equilibrium, a restoring force acts towards that position.

For an ideal spring, the restoring force is proportional to the displacement $$$x$$$ . If $$$k$$$ is the spring constant and $$$m$$$ is the mass, Newton’s second law gives:

F=-kx, \qquad a=\frac{F}{m}=-\frac{k}{m}x

The minus signs show that the force and acceleration point in the opposite direction to the displacement. As the mass passes through equilibrium, the restoring force is zero, but the mass continues moving. On the other side, the force reverses direction and pulls it back.

Because acceleration is proportional to displacement and directed towards equilibrium, the motion is simple harmonic. Its angular frequency $\omega$ and period $$$T$$$ are:

\omega=\sqrt{\frac{k}{m}}, \qquad T=2\pi\sqrt{\frac{m}{k}}

The period is the time for one complete oscillation. A larger mass gives a longer period: at the same displacement, the same restoring force produces a smaller acceleration. A larger spring constant gives a shorter period: the spring produces a greater restoring force at the same displacement.

The amplitude is the maximum displacement from equilibrium. For an ideal spring, changing the amplitude does not change the period. At a larger displacement, the restoring force is proportionally larger, so the mass also accelerates more strongly.

Simple pendulum oscillations

A simple pendulum consists of a mass suspended from a fixed point. When displaced and released, the mass swings back and forth about its lowest position. This lowest point is the equilibrium position: gravity has no component along the pendulum’s path there.

On either side of equilibrium, a component of gravity acts along the curved path towards the lowest point. It gives the mass a restoring acceleration, which changes direction as the mass passes through equilibrium. The mass continues past equilibrium, slows as it rises, and then swings back.

The small-angle approximation

If the angular displacement from equilibrium is $\theta$ , the acceleration along the path is $-g\sin\theta$ , where $$$g$$$ is the gravitational field strength. The minus sign shows that the acceleration is directed towards equilibrium, opposite to the displacement.

For small angles measured in radians, $\sin\theta \approx \theta$ . The acceleration is therefore approximately proportional to $-\theta$ . Since displacement along the arc is also proportional to $\theta$ , the restoring acceleration is approximately proportional to displacement from equilibrium and directed towards it. The pendulum’s motion is then approximately simple harmonic.

Period of a simple pendulum

The period $$$T$$$ is the time for one complete oscillation. For small-angle oscillations, it is given by:

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

Here $$$l$$$ is the pendulum length, measured from the suspension point to the mass, and $$$g$$$ is the gravitational field strength.

A longer pendulum has a longer period: specifically, $$$T$$$ is proportional to $\sqrt{l}$ . The period does not depend on the mass. Within the small-angle approximation, it also does not depend on the amplitude, the maximum displacement from equilibrium.

Deriving periods of harmonic oscillators

To find the period of an oscillator, first determine how its acceleration depends on its displacement from equilibrium. If the acceleration is proportional to the displacement and always points back towards equilibrium, the motion is simple harmonic motion (SHM):

a=-\omega^2x

Here, $$$x$$$ is displacement from equilibrium, $$$a$$$ is acceleration and $\omega$ is angular frequency. The minus sign shows that the acceleration acts in the opposite direction to the displacement.

If the system’s restoring acceleration can be written as $$$a=-kx$$$ , where $$$k$$$ is positive, comparison with the SHM equation gives $\omega^2=k$ . The period is then:

T=\frac{2\pi}{\omega}=\frac{2\pi}{\sqrt{k}}

The quantity to identify as $\omega^2$ is the positive coefficient of displacement. Take its square root to find $\omega$ before calculating the period.

Period of liquid oscillating in a U-tube

Consider liquid of density $\rho$ in a U-tube of uniform cross-sectional area $$$A$$$ . Let the total length of the liquid column be $$$L$$$ , and neglect friction. At equilibrium, the liquid surfaces are at the same height.

If the liquid is displaced so that one surface rises by $$$x$$$ , the other falls by $$$x$$$ . The difference in surface heights is therefore $$$2x$$$ . This produces a pressure difference of $2\rho gx$ and a restoring force of magnitude $2\rho gAx$ , where $$$g$$$ is gravitational acceleration.

The whole liquid column moves, so its mass is $\rho AL$ . Taking the direction of the initial displacement as positive, the restoring acceleration is:

a=\frac{-2\rho gAx}{\rho AL}=-\frac{2g}{L}x

This has the SHM form, with $\omega^2=2g/L$ . Hence the period is:

T=2\pi\sqrt{\frac{L}{2g}}

The density and cross-sectional area cancel: they affect both the restoring force and the mass being accelerated.

Approximate SHM and small angles

Some oscillators have a restoring acceleration involving $\sin\theta$ rather than being exactly proportional to angular displacement $\theta$ . For a small angle measured in radians, the approximation $\sin\theta\approx\theta$ can turn the acceleration into the SHM form.

For example, if angular acceleration is $-C\sin\theta$ , it is approximately $-C\theta$ at small angles. Comparison with the angular form of the SHM equation gives $\omega^2=C$ . The resulting period is an approximation that applies when the angular displacement remains small.

Energy in simple harmonic motion

In undamped simple harmonic motion, energy transfers between kinetic energy and potential energy. No mechanical energy is lost, so their sum remains constant throughout the oscillation.

As the object moves away from equilibrium, it slows down: kinetic energy decreases while potential energy increases. As it returns towards equilibrium, it speeds up and the transfer reverses.

At equilibrium, speed and kinetic energy are maximum, while potential energy is minimum.
At either extreme displacement, the object is momentarily at rest. Its kinetic energy is zero and its potential energy is maximum.

Energy and displacement

Take potential energy to be zero at equilibrium. For an object of mass $$$m$$$ , displacement $$$x$$$ , amplitude $$$A$$$ and angular frequency $\omega$ , the energies are:

\begin{aligned} E_p &= \frac{1}{2}m\omega^2x^2 \\ E_k &= \frac{1}{2}m\omega^2(A^2-x^2) \\ E_{\text{total}} &= \frac{1}{2}m\omega^2A^2 \end{aligned}

Potential energy increases with the square of the displacement magnitude. Kinetic energy falls by the same amount, so the total remains constant. The energies are the same at $$$x$$$ and $$$-x$$$ because both expressions depend on $$$x^2$$$ .

At equilibrium, $$$x=0$$$ , so $$$E_p=0$$$ and all the mechanical energy is kinetic. At either extreme, $x=\pm A$ , so $$$E_k=0$$$ and all the mechanical energy is potential.

Energy changes during an oscillation

If the displacement varies with time as $x=A\cos(\omega t)$ , the energies vary as:

\begin{aligned} E_p &= \frac{1}{2}m\omega^2A^2\cos^2(\omega t) \\ E_k &= \frac{1}{2}m\omega^2A^2\sin^2(\omega t) \end{aligned}

During one complete oscillation, the object reaches two extremes and passes through equilibrium twice. Potential energy therefore reaches its maximum twice, as does kinetic energy. Each energy completes two cycles per oscillation, while their sum remains constant.

Damping of oscillations

An oscillator moves back and forth about an equilibrium position. Damping occurs when its mechanical energy is dissipated to the surroundings, for example by friction or air resistance. As the oscillator loses mechanical energy, it cannot move as far from equilibrium, so its amplitude decreases.

The mechanical energy of the oscillating system decreases as damping transfers energy to the surroundings. Energy is not destroyed: it is dissipated, often as thermal energy.

The effect of damping strength

With light damping, the oscillator continues to pass back and forth through equilibrium, but each successive oscillation has a smaller amplitude. It gradually comes to rest at equilibrium.

With sufficiently heavy damping, the oscillator returns towards equilibrium without oscillating. Increasing the damping beyond this point makes the return slower.

Critical damping is the boundary between oscillatory and non-oscillatory behaviour. It gives the fastest return to equilibrium without oscillation.