Simple Harmonic Motion — The Equations That Matter
Simple harmonic motion (SHM) is one of the few Class 11 chapters where almost every question can be answered with five formulas — provided you know which one to reach for. Students lose marks here not because the physics is hard, but because they mix up amplitude with displacement, invert the time-period formula, or forget that the angle inside the sine is measured in radians. This guide sets out the equations, works four numerical examples with every step shown, and cross-checks two of them by a second route so you can see that the answers are genuinely consistent.
What makes a motion simple harmonic?
A body is in SHM when the restoring force always pulls it back towards a fixed centre and is directly proportional to how far it has strayed from that centre:
The minus sign is the whole idea: the force points opposite to the displacement. If a question hands you an acceleration proportional to displacement with a negative sign, the motion is SHM, and the number multiplying x is ω². That single observation answers a surprising number of one-mark questions.
The five equations you actually need
2. Time period: T = 2π / ω = 2π √(m / k) (s)
3. Frequency: f = 1 / T = ω / 2π (Hz)
4. Motion: x = A sin(ωt + φ), v = ω √(A² − x²), vmax = ωA, amax = ω²A
5. Energy: E = ½ k A² (total), PE = ½ k x², KE = ½ k (A² − x²)
What every symbol means
| Symbol | Meaning | SI unit |
|---|---|---|
| k | Force constant (spring constant) — how stiff the restoring system is | N m⁻¹ |
| m | Mass of the oscillating body | kg |
| ω | Angular frequency — radians swept per second | rad s⁻¹ |
| T | Time period — the time for one complete oscillation | s |
| A | Amplitude — maximum displacement from the centre | m |
| x | Displacement at that instant, measured from the centre | m |
| φ | Phase constant — where in its cycle the motion started | rad |
Worked example 1 — finding ω, T and f
A 0.20 kg block on a spring of force constant k = 80 N m⁻¹ oscillates on a frictionless surface. Find ω, T and f.
ω = √(k / m) = √(80 ÷ 0.20) = √400 = 20 rad s⁻¹
T = 2π / ω = 6.2832 ÷ 20 = 0.3142 s
f = 1 / T = 1 ÷ 0.3142 = 3.183 Hz
Cross-check: f should also equal ω / 2π = 20 ÷ 6.2832 = 3.183 Hz. Two independent routes give the same number, so the working is sound.
Worked example 2 — maximum speed, maximum acceleration and total energy
The same block is pulled 5.0 cm from the centre and released, so A = 0.050 m.
vmax = ωA = 20 × 0.050 = 1.0 m s⁻¹
amax = ω²A = 400 × 0.050 = 20 m s⁻²
E = ½ k A² = 0.5 × 80 × (0.050)² = 0.5 × 80 × 0.0025 = 0.10 J
Cross-check: at the centre the whole energy is kinetic, so E = ½ m vmax² = 0.5 × 0.20 × (1.0)² = 0.10 J. Identical — good.
Worked example 3 — speed at a given displacement
Find the speed of the same block when it is 3.0 cm from the centre, so x = 0.030 m.
A² = 0.0025, x² = 0.0009, A² − x² = 0.0016
v = ω √(A² − x²) = 20 × √0.0016 = 20 × 0.040 = 0.80 m s⁻¹
Cross-check by energy:
PE at that point = ½ k x² = 0.5 × 80 × 0.0009 = 0.036 J
KE = E − PE = 0.10 − 0.036 = 0.064 J
v = √(2 × KE ÷ m) = √(2 × 0.064 ÷ 0.20) = √0.64 = 0.80 m s⁻¹ — the same answer.
Worked example 4 — the simple pendulum
For small angular displacements (roughly under 10°, where sin θ ≈ θ in radians) a pendulum is also in SHM, but with its own period formula:
A pendulum of length L = 1.00 m, taking g = 9.81 m s⁻².
L / g = 1.00 ÷ 9.81 = 0.10194
√0.10194 = 0.31928
T = 2π × 0.31928 = 6.2832 × 0.31928 = 2.006 s ≈ 2.01 s
Notice what is not in the formula: the mass of the bob, and the amplitude. Both are irrelevant for small swings, which is exactly why pendulum clocks work. Many textbooks use g = 9.8 m s⁻² instead of 9.81; with g = 9.8 the same pendulum gives T = 2.007 s. Either value is accepted in exams — use one consistently and say which one you used.
Reading the motion without a graph
A displacement–time graph would show a sine curve, but you do not need the picture to answer questions about it. This table is the graph in words, for one full cycle starting from the centre:
| Position | Displacement x | Speed | Acceleration | Energy split |
|---|---|---|---|---|
| Centre (mean position) | 0 | Maximum, ωA | Zero | All kinetic |
| Halfway out | A/2 | ω√(3)A/2 ≈ 0.87 ωA | ω²A/2 | 25% potential, 75% kinetic |
| Extreme position | ±A | Zero | Maximum, ω²A | All potential |
Check the middle row against example 3 if you like: at x = A/2 the potential energy is ½k(A/2)² = ¼ × ½kA², which is one quarter of the total. The kinetic share is therefore three quarters, and since KE goes as v², the speed is √(3)/2 of the maximum.
Common mistakes that cost marks
- Inverting the period formula. T = 2π√(m/k), not 2π√(k/m). Sanity test: a heavier mass must oscillate more slowly, so m has to be on top.
- Confusing amplitude with total path length. In one full oscillation the body travels 4A — not A and not 2A.
- Working in degrees. In x = A sin(ωt + φ) the quantity ωt is in radians. Put the calculator in radian mode before starting an SHM question.
- Dropping the minus sign in a = −ω²x. The magnitude is ω²x, but the direction is always back towards the centre, and derivation questions are marked on that sign.
- Using the pendulum formula for large swings. T = 2π√(L/g) is a small-angle result; beyond roughly 10° the true period is longer than the formula predicts.
- Mixing centimetres with metres. Convert A and x to metres before using them alongside k in N m⁻¹, or the energy comes out ten thousand times wrong.
- Assuming energy oscillates. The kinetic and potential shares swap continuously, but their sum stays at ½kA² throughout — that constancy is often the fastest route to an answer.
Where SHM appears in exams
| Exam | Typical use |
|---|---|
| CBSE / ICSE Class 11 | Spring–mass and pendulum numericals; deriving v = ω√(A² − x²) |
| JEE / NEET | Energy split at a given displacement; springs in series and parallel |
| Class 12 physics | LC oscillations and AC circuits reuse exactly the same mathematics |
| Chemistry, Class 12 onwards | The harmonic-oscillator model of a vibrating bond, which underlies infrared spectroscopy |
That last row is worth pausing on. A chemical bond behaves, to a first approximation, like a spring joining two masses, so the same ω = √(k/m) idea explains why a C–H stretch appears at a higher wavenumber than a C–Cl stretch: hydrogen is much lighter, so ω is larger. Learning SHM properly in Class 11 pays you back when spectroscopy arrives.
Check the frequency side of your working. Once you have ω, the period–frequency–wavelength arithmetic is quick to verify with the Waves calculator, which handles the f = 1/T and v = fλ relations that SHM and wave questions share.
Open the Waves Calculator →Studying Class 11–12 and finding that physics-style numericals keep turning up in your chemistry paper? ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre and online classes across India — details at abcchemistry.in.