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Recognise SHM by a = −ω²x, Then Read ω, A and the Phase

Simple harmonic motion is defined by one condition: the acceleration is proportional to the displacement and points back toward the mean position, a = -ω² x. Everything else, period, speed, phase and timing, follows from that.

NSEJS · Oscillations · 17 linked questions · 100% coverage

Know this first

Motion and force questions become simpler when direction, starting condition, time and the acting forces are marked first.

Distance · Length of the path travelledDisplacement · Straight-line change in positionResultant force · Net force including direction

Study deeper: NCERT Science · Oscillations

Recognising SHM and reading ω

The equation of motion is d²x/dt² = -(k/m) x = -ω² x. When a question gives x, v and a at an instant, use ω² = a/x: for x = 1 cm, a = 1 cm/s² the period is T = 2π s.

From a + 16π²

x = 0, ω = 4π and T = 0.5 s

From a = 16 m/s² at x = 4 m, ω² = 4 and T = π s

The force constant

k = m \omega ² = 4\pi ² m n²

A function is SHM only if it is a single sinusoid: sin ωt - cos ωt = sqrt(2) sin(ωt - π/4) and 3 cos(3π/4 - 5ωt) are SHM, but sin³ωt (a sum of different frequencies) and 1 + ωt + ω²t² (not periodic) are not.

Speed, acceleration and phase

For x = A sin(ωt + φ): v = Aω cos(ωt + φ) = ω sqrt(A² - x²), v_max = Aω at the mean position and a_max = ω² A at the extremes

A particle of frequency 300 Hz and amplitude 0.1 cm has

v_{max} = 2\pi x 300 x 0.1 = 60\pi cm/s

For two SHMs of the same amplitude with angular frequencies 500 and 5000 rad/s, the maximum accelerations are in the ratio (500/5000)² = 1 : 100.

For x = 5 cos(2πt + π) at t = 1.5 s the phase is 4π, so x = 5, v = 0 and a = -(2π)² x = -20π²: at the end of the swing

A phase of π/6 in

x = a sin(phase) means x = a/2 and moving away from the mean position in the positive direction

Timing and reading graphs

From the mean position to A/2 takes T/12 and from A/2 to A takes T/6, so the ratio of the two times is 1 : 2 (use sin ωt = 1/2 at ωt = 30°, and sin ωt = 1 at 90°).

A velocity-time graph that repeats after 0.04 s has

T = 0.04 s and f = 25 Hz

On a vₓ plot, v_max is at

x = 0, the amplitude is the x-intercept, and ω = v_max / A: for v_max = 0.4 m/s and A = 0.1 m, ω = 4 rad/s and T = π/2 s

Amplitude stays constant for free, maintained and forced vibrations but decays for damped vibrations.

A block on a vibrating piston

A block resting on a piston that oscillates vertically stays in contact as long as the piston's downward acceleration is less than g.

It separates when ω²

A = g

For a period of 2 s (

ω = π), A = g/π², close to 1 m, and the maximum speed of the piston is ω A = g/ω, about 3.14 m/s

OlympIQ exam lens

✓SHM: a = -ω² x; T = 2π/ω; use ω² = a/x from an instant
✓v = ω sqrt(A² - x²); v_max = Aω; a_max = ω² A; k = m ω²
✓From mean to A/2 takes T/12; from A/2 to A takes T/6; mean to A/sqrt2 takes T/8
✓A function is SHM only if it is a single sinusoid
✓A block leaves a vertically vibrating support when ω² A = g

OlympIQ trap

Do not assume any periodic motion is SHM. A function such as sin³ωt repeats but is not a single sinusoid, so its acceleration is not proportional to the displacement.

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