Simple Harmonic Motion

A child on a swing, a guitar string, the air in a flute, atoms in a crystal: all of them move back and forth around a resting point. The simplest and most important of these motions has a name: simple harmonic motion.

Periodic motion and oscillation#

Motion that repeats itself after a fixed interval of time is called periodic, and that interval is the time period. When the body moves to and fro over the same path, the periodic motion is called an oscillation. The Earth going round the Sun is periodic but not an oscillation; a swinging pendulum is both.

What makes it “simple harmonic”#

Put the fixed point (the centre of oscillation) at the origin and the line along x:

In an inertial frame a = F/m, so the same statement in terms of force is:

The net force is zero at x = 0, so the centre of oscillation is the equilibrium position. A force that always pushes back toward equilibrium is called a restoring force; SHM is what a restoring force proportional to displacement produces.

A block on a spring#

Put a block of mass m on a smooth table, attached to a wall by a spring of constant k. At O the spring has its natural length. Pull the block to P, a distance A to the right, and let go. Its displacement x always equals the stretch of the spring, and the spring force is −kx: by definition, this is SHM. Follow one cycle:

• From P to O the force points toward O, so the block speeds up. It is fastest at O.
• Past O the spring is compressed and pushes right, so the block slows, stopping at some point Q.
• Then it is pushed back, speeds up to O, slows again and stops at P. And so on.

The block is momentarily at rest at both P and Q, so all the energy is in the spring at those points. With no friction,

The block swings equal distances either side of the centre. That maximum distance is the amplitude A.

From a = −ω²x alone, a little calculus gives the complete story: where the particle is, how fast it moves, at every moment.

Speed at each position#

Write $a = \dfrac{dv}{dt} = \dfrac{dv}{dx}\dfrac{dx}{dt} = v\dfrac{dv}{dx}$. Then $v\,dv = -\omega^2x\,dx$. Integrate from the starting state (x₀, v₀) to a general (x, v):

Position at each moment#

Now $\dfrac{dx}{\sqrt{A^2 - x^2}} = \omega\,dt$. Integrating, $\sin^{-1}(x/A) = \omega t + \delta$, where δ is a constant fixed by the starting position. So:

The vocabulary of SHM#

Amplitude A

Since sin lies between −1 and +1, x lies between −A and +A: A is the greatest displacement from the centre.

Time period T

One complete oscillation brings the particle back to the same position moving the same way (Q → P → Q, or O → P → Q → O). Both sin and cos repeat when the angle grows by 2π, so $\omega T = 2\pi$:

Frequency ν and angular frequency ω

The frequency is the number of oscillations per second, $\nu = 1/T = \dfrac{\omega}{2\pi} = \dfrac1{2\pi}\sqrt{\dfrac km}$ (12.9, 12.10), measured in hertz (Hz). ω itself is the angular frequency, in rad/s.

Phase and phase constant

The angle $\phi = \omega t + \delta$ is the phase. It tells you the state of the motion: φ = 0 means passing the centre moving in the +x direction; φ = π/2 means at +A, momentarily at rest; φ = π, passing the centre moving in the −x direction; φ = 3π/2, at −A. Adding 2π brings you back to the same state.

δ, the phase constant, depends only on which instant you call t = 0. Start the clock as the particle passes the centre going forward and δ = 0, so x = A sin ωt; start it at the positive extreme and δ = π/2, so x = A cos ωt. The sine and cosine forms describe the same motion with different δ.

Shine a light from the side at a ball going round a turntable and watch its shadow on the wall. The shadow bobs back and forth in perfect simple harmonic motion.

The projection#

Let a point P move round a circle of radius A at constant angular speed ω, starting on the x-axis at t = 0. Take the centre as the origin. At time t the radius OP has turned through ωt, so the feet of the perpendiculars from P onto the axes are at

Each is exactly the equation of SHM, with amplitude A (the circle’s radius) and angular frequency ω (the angular speed). The two shadows are a quarter-cycle apart, because $\cos\omega t = \sin(\omega t + \pi/2)$.

An oscillator is an energy exchange. At the ends of the swing everything is stored; in the middle everything is motion. The total never changes.

Potential energy of a linear restoring force#

The force F = −kx does work $\int_0^x(-kx)\,dx = -\tfrac12kx^2$ as the displacement grows from 0 to x. Potential energy rises by the negative of that work. Choosing U = 0 at the centre:

Kinetic and total energy#

With x = A sin(ωt + δ) and v = Aω cos(ωt + δ):

At the centre U = 0 and all of E is kinetic: $\tfrac12mv_{\max}^2 = \tfrac12m\omega^2A^2$. At x = ±A the body is momentarily still and all of E is potential.

Galileo, so the story goes, timed a swinging cathedral lamp with his pulse and noticed that big and small swings took the same time. The pendulum became the world’s timekeeper for three centuries.

Angular SHM#

A body free to turn about an axis (a hanging umbrella nudged aside, say) can oscillate in angle. We call it angular simple harmonic motion when there is a position θ = 0 of zero torque, and a displacement θ produces a torque proportional to θ that tries to bring the body back:

This is the SHM equation with θ in place of x, so everything carries over:

The simple pendulum#

Let the string, of length l, make angle θ with the vertical. The tension passes through the pivot (no torque); the weight mg has lever arm l sin θ about the pivot and tries to restore θ = 0:

That isn’t proportional to θ, so a pendulum is not exactly SHM. But for small angles $\sin\theta \approx \theta$ (in radians), giving Γ = −mglθ. With $I = ml^2$:

You get the same result treating the bob’s motion along its arc (x = lθ) as linear SHM: the tangential force is −mg sin θ ≈ −mg x/l, so $a = -(g/l)x$.

Measuring g with a pendulum#

Hang a small ball from a clamp by a light thread. The effective length l runs from the point of suspension to the centre of the ball: thread length plus the ball’s radius (measured with slide callipers). Pull the ball slightly aside, release it, time 20 or 50 oscillations with a stopwatch, and divide to get T. Then $g = 4\pi^2l/T^2$. Repeat for several lengths to reduce random errors.

Real swinging things are not point masses on threads. A ring on a nail, a swinging leg, a door closer: their rhythms follow the same rule once we let the moment of inertia take charge.

The physical pendulum#

Any rigid body hung from a fixed pivot O is a physical (or compound) pendulum. It hangs at rest with its centre of mass C directly below O. Let OC = l. Turned through θ, the support force at O has no torque, and the weight has torque

Comparing with 2π√(L/g), it swings like a simple pendulum of length $L = I/(ml)$, the equivalent length. For a point bob, $I = ml^2$ and L = l, as it should.

The torsional pendulum#

Hang a body (a disc, say) from a thin wire and twist it through θ about the wire. The wire twists too, and its elasticity produces a restoring torque proportional to the twist, $\Gamma = -k\theta$, where k is the wire’s torsional constant. This is exactly angular SHM:

What if two restoring forces act on the same particle at once? The answer is pleasantly simple: the motions just add.

Motions add#

Suppose force F₁ alone would move a particle along $\vec r_1(t)$, and F₂ alone along $\vec r_2(t)$. Then $m\ddot{\vec r}_1 = \vec F_1$ and $m\ddot{\vec r}_2 = \vec F_2$. Adding, $m(\ddot{\vec r}_1 + \ddot{\vec r}_2) = \vec F_1 + \vec F_2$, so $\vec r = \vec r_1 + \vec r_2$ is the motion under both forces (with matching starting conditions). Find each motion separately and add the displacements.

Two SHMs along the same line#

Take equal frequencies with a phase difference δ:

Expanding the second and collecting, $x = (A_1 + A_2\cos\delta)\sin\omega t + (A_2\sin\delta)\cos\omega t$, which can be rewritten as a single sine:

The result is another SHM of the same frequency. Its amplitude depends on the phase difference:

• δ = 0 (in phase): $A = A_1 + A_2$, the largest possible.
• δ = π (out of phase): $A = |A_1 - A_2|$; equal amplitudes cancel completely and the particle stays still.
• Anything in between gives an amplitude between these limits.

The phasor (vector) method

Represent each SHM by an arrow whose length is its amplitude, drawn at an angle equal to its phase: A₁ along a reference line, A₂ at angle δ to it. Their vector sum has exactly the magnitude and angle given above. It works for any number of SHMs of the same frequency: draw the arrows head to tail.

Two SHMs at right angles#

Now let one force produce motion along x and the other along y, again at the same frequency:

The particle moves in a plane. Eliminating t gives the path:

• δ = 0: $y = (A_2/A_1)x$, a straight line along one diagonal of the rectangle: SHM of amplitude $\sqrt{A_1^2 + A_2^2}$.
• δ = π: a straight line along the other diagonal.
• δ = π/2: $x^2/A_1^2 + y^2/A_2^2 = 1$, an ellipse with axes along x and y. With A₁ = A₂ it is a circle: two equal perpendicular SHMs a quarter-cycle apart make uniform circular motion (clockwise or anticlockwise depending on which leads).

A real swing doesn’t go on forever; it dies away unless someone keeps pushing. And if the pushes come at just the right rhythm, even small ones build a huge swing.

Damping#

Friction at joints, air resistance and many other effects add a damping force that opposes the velocity and steadily removes energy. In many practical cases it is proportional to the speed, $F = -bv$, and Newton’s second law becomes

For light damping the solution is an oscillation whose amplitude shrinks exponentially:

So the system swings at almost its natural frequency while its amplitude $A = A_0e^{-bt/2m}$ decays. With enough damping it doesn’t oscillate at all: displaced, it simply returns to equilibrium without overshooting. The smallest damping that does this is critical damping, and it brings the system to rest fastest.

Forced oscillations#

Now add a periodic driving force $F_0\sin\omega t$, as when you push a swing:

After an initial messy period, the body settles into steady oscillation at the driving frequency ω (not its natural one), with the energy lost to damping supplied by the driving force. The steady amplitude is

Resonance#

Vary the driving frequency and the amplitude peaks at $\omega = \sqrt{\omega_0^2 - b^2/2m^2}$, very close to ω₀ when damping is small. This is resonance: drive a system at its own natural frequency and it responds enormously. The smaller the damping, the taller and sharper the peak; with no damping at all the formula would give infinite amplitude.