Simple Harmonic Motion Field Guide

01A mass on a spring: force and acceleration

Pull the block to +x₀ and press Release. The spring force F = −kx always points back to the equilibrium position, and it is largest at the ends. Newton’s second law gives a = F/m = −(k/m)x, so the acceleration arrow always points the same way as the force. The velocity is largest in the middle and zero at the ends. With k = 40 N m⁻¹ and m = 1.0 kg: ω² = k/m = 40 s⁻², T = 0.99 s, and at x = 5.0 cm the force is −2.0 N.

02The conditions for SHM: test the force

SHM needs two things: the restoring force (so the acceleration) is proportional to the displacement, and it is directed towards the equilibrium position. Choose a force and move the block. For the spring (k = 20 N m⁻¹) at x = 6.0 cm the force is −1.2 N, and doubling x doubles F: both conditions hold. A ball in a V-shaped groove feels a force of the same size at every x, and a stiff spring with F = −cx³ gives 8 times the force for twice the displacement: these oscillate, but they are not SHM. A force away from the centre gives no oscillation at all. Press Let go to see each motion.

03The simple pendulum: only for small angles

The restoring force is the part of the weight along the arc, mg sin θ. For a small angle in radians sin θ ≈ θ, so the force is proportional to the displacement and the motion is SHM with T = 2π√(l/g): 2.01 s for l = 1.00 m. The dashed bob follows the SHM model and the solid bob is the real pendulum. At 10° the real period is only 0.2 % longer; at 60° it is 7.3 % longer, and the two bobs drift apart. Change the mass: the period does not change.

04The a–x graph: gradient = −ω²

a = −ω²x is a straight line through the origin with a negative gradient, and the gradient is −ω². For a mass on a spring ω² = k/m: with k = 40 N m⁻¹ and m = 1.0 kg the gradient is −40 s⁻², so ω = √40 = 6.32 rad s⁻¹ and T = 0.99 s. Double the mass and the gradient halves to −20 s⁻². The line stops at ±x₀ (here a_max = 40 × 0.05 = 2.0 m s⁻²), because the object never goes further. For any other system the method is the same: ω² = −gradient.

05Displacement, velocity and acceleration against time

Slide the time cursor (x₀ = 4.0 cm, T = 2.0 s, so ω = π = 3.14 rad s⁻¹). The velocity is the gradient of the x–t graph: zero at a peak, largest where x crosses zero, with v_max = ωx₀ = 12.6 cm s⁻¹. The acceleration graph is the x–t graph upside down (a = −ω²x), with a_max = ω²x₀ = 39.5 cm s⁻². At HL, each equation comes from differentiating the one above it. Choose “start at an end” to see the cosine version.

06One cycle, step by step

The block is released from rest at +x₀. Step through one period in eighths. Each row keeps the earlier positions so you can see the whole cycle: the velocity arrow (amber) is longest in the middle and zero at the ends; the acceleration arrow (green) is longest at the ends and zero in the middle, and it always points to the centre. At t = T/8 the block is at 0.71 x₀ and has 71 % of its maximum speed. Moving towards the centre it speeds up; moving away it slows down.

07Energy against displacement and against time

Energy moves back and forth between kinetic and potential, and the total stays the same. Against displacement the two graphs are parabolas that always add up to the total. Against time each energy repeats every T/2, because the block passes the centre twice per period. Here m = 0.50 kg and ω = 10 rad s⁻¹ (T = 0.63 s, so the energy repeats every 0.31 s). HL: E_T = ½mω²x₀² = 1.0 J for x₀ = 0.20 m; halve the amplitude and the total falls to 0.25 J, a quarter. The kinetic and potential energies are equal (0.50 J each) at x = x₀/√2 = 0.14 m, not at x₀/2.

08The phase angle φ: where the clock starts

A point goes round a circle at ω radians per second. Its height moves up and down in SHM: x = x₀ sin(angle). If the point is already at angle φ when the clock starts, the angle at time t is ωt + φ, so x = x₀ sin(ωt + φ). Here x₀ = 0.20 m and T = 2.0 s. With φ = 40° = 0.70 rad the object starts at x = 0.20 sin 40° = 0.13 m, moving in the + direction, 0.22 s ahead of the φ = 0 curve (dashed). Try φ = π/2 (starts at +x₀), π (centre, moving −) and 3π/2 (starts at −x₀). φ never changes the amplitude or the period.

09Speed at any displacement: v = ±ω√(x₀² − x²)

The velocity–displacement graph of SHM is an ellipse. At the centre the speed is largest, v_max = ωx₀; at the ends it is zero. At each x there are two velocities, + and −, because the object passes each point going both ways. With x₀ = 6.0 cm and T = 1.5 s, ω = 4.19 rad s⁻¹ and at x = 3.0 cm, v = ±21.8 cm s⁻¹. That is 0.87 of v_max (25.1 cm s⁻¹), not half of it. Move x past x₀ to see that the object never gets there.

10The vertical spring: SHM about a lower equilibrium

Hang the mass and it rests lower, where ke = mg. For m = 0.20 kg and k = 8.0 N m⁻¹ the extension is e = 0.245 m. Measured from this new equilibrium the extra force is still −kx, so the motion is SHM with the same T = 2π√(m/k) = 0.99 s. Pull it down 0.10 m and release: v_max = ωx₀ = 0.63 m s⁻¹. Watch the bars: the oscillation energy ½kx₀² = 0.040 J is not the elastic energy in the spring, which at the bottom is ½k(e + x₀)² = 0.48 J. Gravitational energy changes too, and the total stays constant.

11Practice and self-check

Before the exam, can you…

• state the two conditions for SHM in words (not only a = −ω²x)?
• explain the minus sign in a = −ω²x?
• derive ω² = k/m from Hooke’s law, and use it only for a mass on a spring?
• explain why a pendulum is SHM only for small angles, and why the mass does not matter?
• find ω from the gradient of an a–x graph (gradient = −ω²)?
• say where v and a are zero and where they are largest?
• describe the energy changes, and read the period from an energy graph (energy repeats every T/2)?
• (HL) differentiate x = x₀ sin ωt to get v and a, with the calculator in radians?
• (HL) find φ from x(0) and the direction of motion, and use Δφ = (Δt/T) × 2π?
• (HL) use v = ±ω√(x₀² − x²) and E_T = ½mω²x₀², and explain the vertical spring?