Simple Harmonic Motion Formulas: Pick the Right One Fast

Every SHM formula comes from two facts: the displacement is x = A cos(ωt + φ), and the restoring force is F = −kx, which always pushes the object back towards the mean position (the equilibrium point, where the restoring force is zero). Learn those two, then pick from the table by what the question gives you.

The core set

• Displacement: x = A cos(ωt + φ)
• Velocity: v = −Aω sin(ωt + φ)
• Acceleration: a = −ω²x
• Spring: F = −kx, so ω = √(k/m)
• Period and frequency: T = 2π/ω = 1/f, so ω = 2πf
• Energy: K = ½k(A² − x²), U = ½kx², E = ½kA² = ½mω²A²

Here A is the amplitude (the largest displacement), ω the angular frequency in rad/s, f the frequency in hertz, T the period in seconds and k the spring constant in N/m. The chapter on Oscillations in the National Council of Educational Research and Training (NCERT) Class 11 Physics textbook writes the displacement as Eq. 13.4 and the total energy as Eq. 13.18.

Which formula for which question

Convert to SI (metric) units, find ω from T, and every other quantity follows from it. With A = 0.05 m and T = 0.4 s, ω = 15.71 rad/s, the maximum speed is 0.785 m/s and the maximum acceleration is 12.34 m/s².

1. ω = 2π/T = 2π/0.4 = 15.71 rad/s.
2. v_max = ωA = 15.71 × 0.05 = 0.785 m/s, at the mean position.
3. a_max = ω²A = 246.7 × 0.05 = 12.34 m/s², at x = ±A.
4. Speed at x = 3 cm: v = ω√(A² − x²) = 15.71 × √(0.05² − 0.03²) = 15.71 × 0.04 = 0.628 m/s.

Run a sanity check on step 4. The numbers 3, 4 and 5 form a right-angled triangle, so √(A² − x²) is exactly 4 cm and v is 0.8 × v_max. Pass: the speed is at most ωA (equal only at the mean position). Fail: a speed above ωA means you have used f instead of ω or forgotten to convert centimetres.

The phase constant φ is the value of the phase (ωt + φ) at t = 0, so it records where in its cycle the motion starts. It is not the initial position, but once you know A, cos φ = x₀/A gives the position. The NCERT chapter says φ can be determined from the displacement at t = 0 when A is known.

Position alone leaves two possible angles, so use the direction as well. Differentiating x = A cos(ωt + φ) gives v₀ = −Aω sin φ, so sin φ has the opposite sign to v₀.

The Maharashtra Std 12 textbook writes the same motion as x = A sin(ωt + α), where α is the initial phase. Since sin θ = cos(θ − π/2), the two forms are linked by φ = α − π/2, or α = φ + π/2. The motion is identical; only the label on the starting angle changes. Pick one form and stay with it through the whole question.

From a graph, read x at t = 0 for cos φ, then look at whether the curve is rising or falling at t = 0 for the sign of sin φ.

A particle moving in a circle of radius A at constant angular speed ω casts a shadow on a diameter, and that shadow performs SHM with amplitude A and angular frequency ω. The angle of the rotating radius at t = 0 is the phase constant φ.

This gives a quick check: the circular speed ωA is exactly the maximum speed v_max of the shadow, and the circular acceleration ω²A is a_max.

Angular SHM follows the same pattern with torque in place of force. A restoring torque τ = −Cθ gives Iα = −Cθ, so ω = √(C/I), θ = θ₀ cos(ωt + φ) and T = 2π√(I/C). To translate any linear result, swap x for θ, k for C and m for I.

The period of a mass on a spring depends only on m and k, T = 2π√(m/k), so any change to a spring is a change to k. Cutting a spring in half doubles k, because each half is stiffer, and T becomes 1/√2 = 0.707 of the original.

The frequency of the series pair is half that of the parallel pair, a ratio of 1 : 2. For unequal springs, parallel gives k = k₁ + k₂ and series gives 1/k = 1/k₁ + 1/k₂.

For a spring hung vertically, gravity only moves the equilibrium position down by mg/k. Measure x from that new equilibrium and T = 2π√(m/k) stays exactly as it was; g does not appear in it.

The formula is a small-angle result: the derivation replaces sin θ with θ, which holds below about 15°. At 15° the exact period is only 0.43% longer than the formula gives. At 60° it is 7.3% longer and at 90° it is 18.0% longer. OpenStax gives the same 15° limit, and NCERT makes the sin θ ≈ θ step explicit.

Our calculation, October 2026. The exact period is T = 4√(L/g)·K(k), where k = sin(θ₀/2) and K is the complete elliptic integral of the first kind. We evaluated K with the arithmetic-geometric mean and divided by 2π√(L/g), which gives (2/π)K(k).

Use T = 2π√(L/g) whenever a question says "small oscillations", and do not expect it to be exact beyond that. The mass of the bob and the amplitude do not appear in it. L is measured to the centre of the bob. A 1 m pendulum has T = 2π√(1/9.8) ≈ 2.01 s. In a lift accelerating upwards at a, replace g with g + a.

Total energy stays fixed at E = ½kA² = ½mω²A², while kinetic and potential energy swap. At the mean position all of it is kinetic, and at x = ±A all of it is potential. NCERT notes that kinetic and potential energy both repeat after T/2, so each peaks twice per cycle.

• Kinetic energy at x: K = ½k(A² − x²). Potential energy: U = ½kx².
• Kinetic and potential energy are equal where x = ±A/√2, about 0.707A.
• At x = A/2, K = ¾E and U = ¼E.
• Rewriting E with ω = 2πf gives E = 2π²mf²A². Doubling A alone multiplies E by 4. Doubling A and halving f multiplies E by 4 × ¼ = 1, so E is unchanged.

Practise by question type with one fixed routine, so the choice of formula becomes automatic. We recommend three steps: write down what is given in SI units, find ω first, then test the answer against the ceilings v ≤ ωA and a ≤ ω²A. In Maharashtra, SHM sits in the Higher Secondary Certificate (HSC) Standard (Std) 12 Oscillations chapter, and NCERT places it in Class 11 Chapter 13.

Drill each row of the table in turn: speed at x, a spring change, a pendulum, an energy ratio, a phase from a graph. Our plan for MHT-CET Physics previous-year questions (PYQs) shows how to organise past papers chapter by chapter, and our HSC Physics formula sheet keeps the Maharashtra formulas in one place.

RG Lectures, founded by Rahul Jewaani (RG Sir) in May 2018, teaches Physics online. Padhna Aasaan Hai. We post free lectures on our YouTube channel, and Class 11 students can start with our MHT-CET 2028 Class 11 Physics batch. For MHT-CET 2027, our Long Term Batch is ₹1,632 as of October 2026.