Oscillations — Formula Sheet

$F$: Restoring force (N) · $k$: Force constant (N m⁻¹) · $x$: Displacement from the mean position (m)

The restoring force is proportional to the displacement from the mean position and acts towards it (hence the minus sign). k is the force constant, unit N m⁻¹.

$\omega$: Angular frequency (rad s⁻¹) · $m$: Mass of the oscillating particle (kg)

Any system whose equation of motion reduces to this form performs SHM with angular frequency ω. Unit of ω: rad s⁻¹.

$A$: Amplitude (m) · $\varphi$: Initial phase or epoch (rad) · $t$: Time (s)

A is the amplitude (maximum displacement). (ωt + φ) is the phase and φ is the initial phase (epoch). If the particle starts at the mean position moving in the positive direction, φ = 0; if it starts at the positive extreme, φ = π/2.

$v$: Velocity at displacement x (m s⁻¹) · $v_{max}$: Maximum speed (m s⁻¹)

Maximum at the mean position (x = 0) and zero at the extreme positions (x = ±A).

$a$: Acceleration at displacement x (m s⁻²) · $a_{max}$: Maximum acceleration (m s⁻²)

Always directed towards the mean position. Zero at the mean position and maximum (Aω²) at the extremes. Dividing a_max by v_max gives ω.

$T$: Period (s) · $n$: Frequency (Hz) · $m$: Mass (kg) · $k$: Force constant (N m⁻¹)

Useful general form: T = 2π√(displacement/acceleration), using magnitudes. The period does not depend on the amplitude.

$R$: Resultant amplitude (m) · $A_1,A_2$: Amplitudes of the two SHMs (m) · ${\varphi }_1,{\varphi }_2$: Initial phases of the two SHMs (rad)

For two SHMs of the same period along the same path: x₁ = A₁ sin(ωt + φ₁) and x₂ = A₂ sin(ωt + φ₂). The resultant is also an SHM of the same period. In phase: R = A₁ + A₂. Opposite phase: R = |A₁ − A₂|.

$\delta$: Initial phase of the resultant SHM (rad)

Comes from dividing R sin δ by R cos δ. Check the signs of the numerator and denominator to place δ in the correct quadrant.

$E_K$: Kinetic energy (J)

Maximum (½kA²) at the mean position, zero at the extremes.

$E_P$: Potential energy (J)

Equal to the work done against the restoring force in displacing the particle from 0 to x. Zero at the mean position, maximum at the extremes.

$E$: Total mechanical energy (J) · $n$: Frequency (Hz)

Constant throughout the motion, independent of x. Proportional to the square of the amplitude and the square of the frequency.

$L$: Length of the pendulum (m) · $g$: Acceleration due to gravity (m s⁻²)

Valid for small amplitudes. Independent of the mass of the bob and of the amplitude. L is measured from the point of suspension to the centre of the bob.

A pendulum with a period of 2 s, so each swing from one extreme to the other takes 1 s. Its length depends on the local value of g (here g = 9.8 m s⁻²).

$\tau$: Restoring torque (N m) · $c$: Torque per unit angular displacement (N m rad⁻¹) · $\theta$: Angular displacement (rad) · $I$: Moment of inertia (kg m²)

The restoring torque is proportional to the angular displacement. c is the torque per unit angular displacement (N m rad⁻¹). Compare with linear SHM: m ↔ I, k ↔ c.

$I$: Moment of inertia of the magnet about the axis of rotation (kg m²) · $\mu$: Magnetic dipole moment of the magnet (A m²) · $B$: Magnetic field (T)

For small angular displacements, the restoring torque is μB sin θ ≈ μBθ, so c = μB. A stronger field or a stronger magnet gives faster oscillations.

$A_0$: Initial amplitude (m) · $b$: Damping constant (kg s⁻¹) · ${\omega }^{\prime }$: Angular frequency of damped oscillation (rad s⁻¹)

A damping force −bv makes the amplitude decay exponentially, and the oscillation is slightly slower than the undamped one (ω' < ω). With b = 0 you get back ideal SHM.