Simple harmonic motion

Start with a moving body and a spring

A 3 kg body is attached to a horizontal spring. Choose a fixed coordinate x that increases to the right. The equilibrium position is x_eq = 1 m. At t = 0 s the body is at x = 0.7 m and already moving right at 0.8 m/s. It begins to the left of equilibrium; being on that side does not mean it is moving left.

Supply an ideal model: a fixed support, a massless linear spring of stiffness k = 12 N/m, and no horizontal friction, damping or external drive. Weight and the normal force cancel, with zero vertical acceleration. These conditions define the example; they are not measurements of a particular spring.

Position, displacement, velocity, and acceleration prepares the signed quantities used here. Geometry, trigonometry, and coordinate systems prepares the sine and cosine graphs. The needed ideas are introduced below, so a first reading can follow the concrete state and the table before returning to the formulas.

Newton’s laws and free-body diagrams and Friction, tension, drag, springs, and contact forces are published return paths for the force model. Derivatives, Taylor series, and local approximation and Ordinary differential equations are optional return paths for checking the motion equation. A dot below means a derivative with respect to actual time t.

Measure from equilibrium

Define ξ = x − x_eq, the signed displacement from equilibrium. Initially ξ = −0.3 m although the absolute position x = 0.7 m is positive. Since x_eq is fixed, velocity v = dξ/dt = dx/dt and acceleration a = d²ξ/dt² = d²x/dt².

The horizontal net force is F = −kξ. A positive displacement gives a leftward force; a negative displacement gives a rightward force. Initially F = −(12 N/m)(−0.3 m) = +3.6 N, so a = F/m = +1.2 m/s².

Newton’s second law gives mξ̈ = −kξ. Dividing by the positive mass defines ω² = k/m = 4 s⁻², where ω = 2 s⁻¹ is the positive natural angular frequency. The equation is ξ̈ = −ω²ξ: acceleration is proportional to displacement and has the opposite sign.

Simple harmonic motion is this sinusoidal motion under a constant positive restoring coefficient. Here the linear force law is stipulated for every real ξ and the mathematical solution is defined for all t ≥ 0 s. A real spring has a limited range of linear behavior; its range is not supplied by this example.

Use both parts of the initial state

A position alone does not select the motion. The body might be moving right, moving left or momentarily at rest there. For this equation write ξ(t) = C cos(ωt) + D sin(ωt), where C and D have units of metres.

At t = 0 s, cosine is 1 and sine is 0, so C = ξ(0) = −3/10 m. Differentiating gives v(0) = ωD; hence D = v(0)/ω = 2/5 m. Both initial data have now been used.

Let τ = t/(1 s), a dimensionless numerical time. The selected solution is xi(t) = (-3/10 m)*cos(2*tau) + (2/5 m)*sin(2*tau). Its absolute position is x(t) = 1 m + ξ(t). The trigonometric argument 2τ is a dimensionless phase expressed in radians.

Differentiating with respect to actual time gives v(t) = (3/5 m/s) sin(2τ) + (4/5 m/s) cos(2τ), then a(t) = (6/5 m/s²) cos(2τ) − (8/5 m/s²) sin(2τ). These satisfy the stated initial data and a(t) = −(4 s⁻²)ξ(t) at every time. The constant-coefficient equation and both initial data select this unique classical solution.

Find amplitude, phase and period

The two sine and cosine coefficients combine into an amplitude A = √(C²+D²) = √(9/100+16/100) m = 1/2 m. Amplitude is the greatest magnitude of equilibrium displacement, not the initial displacement and not the distance traveled in a cycle. The absolute position stays between 0.5 m and 1.5 m.

Write the same solution as ξ(t) = (1/2 m) cos(2τ−φ). Expanding the cosine requires cos φ = −3/5 and sin φ = 4/5. Choose φ in the second quadrant, π/2 < φ < π. The notation φ = atan2(4,−3) means that angle with both of these signs; φ ≈ 2.21429744 rad. An ordinary arctangent of one ratio alone would not fix the correct quadrant.

One complete nonconstant cycle requires the phase to increase by 2π. Thus T = 2π/ω = π s and f = 1/T = 1/π Hz. Angular frequency ω counts radians of phase per second, while frequency f counts cycles per second; ω = 2πf.

The period is independent of amplitude within this ideal linear model. This does not establish amplitude independence for every real oscillator. The constant solution ξ = 0 is a separate case: it has no smallest positive fundamental period, although the equation still has the natural-frequency parameter ω.

Compare three quantities at common times

The three panels share the same actual time axis. Blue circles mark ξ, violet squares mark v and orange diamonds mark a at five quarter-period times. Each curve is the continuous analytic solution; its sampled line segments only approximate the drawing. The markers are chosen model values, not measurements.

The dotted guides mark the two times when velocity is zero. At those times displacement has an extremum and acceleration is nonzero. Conversely, when ξ = 0 the body crosses equilibrium with nonzero speed. The acceleration curve has the opposite sign from the displacement curve throughout.

Read each ordinate using its own units. The vertical scales are 180 pixels per metre for ξ, 90 pixels per m/s for v, and 45 pixels per m/s² for a. Equal heights across these panels are not equal physical quantities. The picture shows one period, 0 ≤ t ≤ π s, within the mathematical domain t ≥ 0 s.

The first five rows give the marker values exactly. The last two give the turning states, using the phase φ defined above. Those two rows are listed separately rather than inserted into chronological order; their times are approximately 1.10714872 s and 2.67794504 s.

Separate turning from equilibrium

In the phase form, v(t) = −(1 m/s) sin(2τ−φ). Its first two positive zeros occur when the phase is 0 and π. The first is t = (φ/2) s: ξ = +0.5 m and x = 1.5 m, but a = −2 m/s² and F = −6 N. The body reverses from rightward to leftward motion.

The second turning time is t = ((φ+π)/2) s. Here ξ = −0.5 m, x = 0.5 m, a = +2 m/s² and F = +6 N. Instantaneous rest does not make either turning state an equilibrium state.

The equilibrium crossings occur at t = ((φ−π/2)/2) s and t = ((φ+π/2)/2) s. Both have ξ = 0, F = 0 and a = 0; their velocities are +1 m/s and −1 m/s. Zero acceleration at an instant does not imply zero velocity or permanently uniform motion.

Over one period the body returns to its initial position and velocity. Its signed displacement is zero, while its traveled distance is 4A = 2 m. The path includes reversals; adding signed endpoint changes would erase that distance.

Check the energy balance

Define kinetic energy K = mv²/2 and choose spring potential energy U = kξ²/2, with U = 0 at equilibrium. Both have units of joules. In the stated fixed-support model there is no damping or drive to change their sum.

Initially K = (3 kg)(0.8 m/s)²/2 = 24/25 J and U = (12 N/m)(0.3 m)²/2 = 27/50 J. Their sum is E = 3/2 J. At either turning point all this mechanical energy is spring potential energy; at an equilibrium crossing it is all kinetic energy.

For this solution K = (3/2 J) sin²(2τ−φ) and U = (3/2 J) cos²(2τ−φ), so their sum stays 3/2 J. A second check is dE/dt = v(ma+kξ) = 0 W. This uses the actual force equation, not an assumption that any periodic motion conserves mechanical energy.

Holding k fixed while changing the mass changes the natural frequency. Holding the initial velocity fixed at the same time also changes the kinetic energy and can change the amplitude. The worked mass comparison below keeps those conditions explicit.

Recognize what the model does not say

Repetition alone is not enough for simple harmonic motion. For example, prescribe the periodic displacement ξ(t) = (1/5 m) cos(2τ) + (1/10 m) cos(4τ), without imposing the spring law. Its period is π s, but its acceleration is −(4/5 m/s²) cos(2τ) − (8/5 m/s²) cos(4τ).

At t = 0 s this prescribed motion has a/ξ = −8 s⁻². At t = (π/4) s it has a/ξ = −16 s⁻². Both displacements are nonzero, so these unequal ratios rule out any single constant restoring coefficient. It is a periodic sum of two frequencies, not one simple harmonic motion.

The main example is exact under its supplied assumptions. A material spring, large extension, moving support, damping or drive needs a model that includes those features. A finite drawing or a few apparently repeating measurements cannot establish the ideal force law.

For a first reading, keep three relationships in view: restoring acceleration opposes equilibrium displacement, both initial data select the phase, and the zero-velocity turns differ from the zero-force crossings. The optional derivative and differential-equation return paths explain why these relationships hold for the whole solution.

Worked examples

Try it yourself

1. For the main initial state, find the net horizontal force and acceleration. Why are they positive even though ξ is negative?

   Show the answer and reasoning

   F = +18/5 N and a = +6/5 m/s².

   F = −kξ reverses the displacement sign. The body is left of equilibrium, so the restoring force and acceleration point right; neither sign is determined by velocity alone.

2. For the main motion at t = (π/4) s, find x, ξ, v and a. Is the body moving toward equilibrium?

   Show the answer and reasoning

   x = 7/5 m, ξ = 2/5 m, v = +3/5 m/s and a = −8/5 m/s². It is moving away from equilibrium at this instant.

   The body is on the right and moving right, although acceleration points left. A restoring force can slow motion away from equilibrium before reversing it.

3. At the main motion’s first positive turning time, which of v, a and F are zero? Give the absolute position.

   Show the answer and reasoning

   Only v is zero. x = 3/2 m, a = −2 m/s² and F = −6 N.

   At phase 0 the displacement is +A. A zero velocity at a turn does not mean zero restoring force.

4. At the main motion’s first equilibrium crossing, give x, v, a and F. Does zero force mean that the body remains there?

   Show the answer and reasoning

   x = 1 m, v = +1 m/s, a = 0 m/s² and F = 0 N. It does not remain there.

   At phase −π/2 the displacement is zero but velocity is positive and maximal. A permanently stationary equilibrium would also require zero velocity.

5. For the two-frequency displacement in the last section, compare a/ξ at 0 s and (π/4) s. Can it satisfy a = −ω²ξ with one constant ω?

   Show the answer and reasoning

   The ratios are −8 s⁻² and −16 s⁻². No single constant ω can fit both.

   At the first time ξ = 0.3 m and a = −2.4 m/s². At the second ξ = −0.1 m and a = +1.6 m/s². Periodicity does not make their unequal ratios equal.

6. For m = 3 kg and k = 12 N/m, set ξ(0) = 0 m and v(0) = 0 m/s. What is the motion, and is π s its smallest positive period?

   Show the answer and reasoning

   ξ, v, a and F stay zero, x stays 1 m, and there is no smallest positive fundamental period.

   The constant solution repeats after every positive duration. The parameter ω = 2 s⁻¹ describes the equation’s nonconstant family, but does not assign a fundamental oscillation period to this stationary state.

What to carry forward

• In the ideal linear model, restoring acceleration opposes displacement from equilibrium, while both initial data select the motion.
• Zero velocity at a turn and zero force at an equilibrium crossing describe different states.
• Amplitude, period and conserved energy follow the stated model; periodicity alone does not establish simple harmonic motion.