Simple Harmonic Motion

SHM is the template for every oscillating system. Advanced tests it through compound/physical pendulums, springs in series/parallel, superposition of SHMs (Lissajous, beats), and identifying SHM in disguised systems (floating blocks, liquid columns, magnet in a coil). The recurring trick is to show acceleration = −ω²x, then read off ω.

Key results

• Defining relation: a = −ω²x; x = A sin(ωt + φ); v = ω√(A²−x²); v_max = ωA; a_max = ω²A.
• Energy: E = ½mω²A²; KE = ½mω²(A²−x²), PE = ½mω²x². KE = PE at x = A/√2.
• Springs: T = 2π√(m/k). Series: 1/k_eff = 1/k1+1/k2 (softer). Parallel: k_eff = k1+k2 (stiffer).
• Simple pendulum: T = 2π√(L/g). Physical pendulum: T = 2π√(I/mgd).
• Superposition: two perpendicular SHMs give ellipses (Lissajous); two along a line with close frequencies give beats.

System	Time period
Spring-mass	2π√(m/k)
Simple pendulum	2π√(L/g)
Physical pendulum	2π√(I/mgd)
Liquid in U-tube (length L)	2π√(L/2g)

Exam Tricks & Tips

• 🎯 Prove SHM by showing a = −ω²x; the coefficient of x gives ω² directly — no need to solve the differential equation.
• 🎯 Series springs are softer, parallel are stiffer — combine k the way you combine capacitors (parallel adds), the opposite of resistors.
• 🎯 KE = PE at x = A/√2, and average KE over a cycle equals average PE, each ¼mω²A².
• 🎯 Phase difference from a reference: velocity leads displacement by 90°, acceleration leads by 180°.
• 🎯 For a floating block or liquid column, restoring force ∝ displacement through buoyancy/weight of extra column — derive ω from that.
• ❌ Common mistake: adding amplitudes of two SHMs directly; you must add them as phasors, A_net = √(A1²+A2²+2A1A2cosΔφ).

Expected exam pattern

1–2 questions: a spring-combination or physical-pendulum time-period numerical, plus a conceptual on energy/phase or an SHM-in-disguise system. Often linked to gravitation (tunnel) or fluids (floating body).

Quick recap

Everything follows from a = −ω²x. Memorise the standard periods, combine springs like capacitors, and use phasor addition for superposed SHMs. Energy splits equally between KE and PE on average; they are equal at x = A/√2.

Cover the answer, recall, then check. 12 cards on SHM for JEE Advanced.

Q1. Defining equation of SHM.
A1. a = −ω²x; acceleration is proportional to and opposite the displacement.

Q2. Velocity of a particle in SHM at displacement x.
A2. v = ω√(A² − x²); maximum ωA at the mean position.

Q3. Time period of a spring-mass system.
A3. T = 2π√(m/k).

Q4. Effective constant for two springs in parallel and in series.
A4. Parallel: k1 + k2 (stiffer). Series: 1/k = 1/k1 + 1/k2 (softer).

Q5. At what displacement is KE equal to PE?
A5. x = A/√2.

Q6. Total energy of SHM.
A6. E = ½mω²A², constant throughout the motion.

Q7. Time period of a physical (compound) pendulum.
A7. T = 2π√(I/mgd), d = distance from pivot to COM.

Q8. Phase relation between displacement, velocity, and acceleration.
A8. Velocity leads displacement by 90°; acceleration leads by 180°.

Q9. How do two perpendicular SHMs of equal frequency combine?
A9. Into a Lissajous figure (an ellipse in general; a line or circle for special phase differences).

Q10. Resultant amplitude of two collinear SHMs (same ω) with phase difference Δφ.
A10. A = √(A1² + A2² + 2A1A2 cosΔφ).

Q11. Time period of a liquid column of total length L oscillating in a U-tube.
A11. T = 2π√(L/2g).

Q12. Fastest way to find ω for a disguised SHM system.
A12. Write the restoring acceleration as −(coefficient)·x; ω² equals that coefficient.

SHM is the template for every oscillation in physics — pendulums, springs, LC circuits, even molecular vibrations. JEE Advanced tests whether you can prove a motion is SHM (not just recognise the spring–mass case), find the effective spring constant of clever arrangements, and handle superposition.

Core concept: a system executes SHM when the restoring "force" is proportional and opposite to displacement: a = −ω²x. Then everything — period, energy, phase — follows from ω.

Core ideas

Beginner — the defining equation

If a = −ω²x, the motion is x = A sin(ωt + φ), with angular frequency ω, period T = 2π/ω, and amplitude A set by initial conditions. Velocity v = ω√(A² − x²) (max at centre, zero at extremes); acceleration a = −ω²x (max at extremes). For a spring–mass, ω = √(k/m); for a simple pendulum (small angle), ω = √(g/L).

Intermediate — proving SHM and effective k

To test any system, displace it slightly, find the net restoring force, and check if it is −(constant)·x. Springs in series give 1/k_eff = 1/k₁ + 1/k₂; in parallel, k_eff = k₁ + k₂. A floating block pushed down feels a buoyant restoring force → SHM with k_eff = ρ_liquid A g. A liquid column of length L in a U-tube oscillates with ω = √(2g/L). Recognising the "effective k" is the whole game.

Advanced — energy, phase, and superposition

Energy shuttles between KE and PE: total E = ½kA² = ½mω²A², constant. KE = ½k(A²−x²), PE = ½kx²; both average to E/2 over a cycle, and the PE curve is a parabola in x but the time-average splits equally. Superposition: two SHMs of the same frequency add to another SHM (phasor addition), with resultant amplitude √(A₁² + A₂² + 2A₁A₂cosδ). Perpendicular SHMs of commensurate frequencies trace Lissajous figures. A physical pendulum has T = 2π√(I/mgd), where d is the pivot-to-CM distance — reducing to the simple pendulum when I = mL².

Worked example

A block of mass m sits on a piston oscillating vertically in SHM with amplitude A and angular frequency ω. At what ω does the block just leave the piston? The block loses contact when the required downward acceleration exceeds g — i.e. at the top extreme where a = ω²A (downward). Contact is lost when ω²A = g, so ω_max = √(g/A). Beyond this the normal force would need to be negative, which is impossible. This "loss of contact when N = 0" condition is a recurring JEE motif.

How JEE Advanced tests this

"Prove this is SHM and find T" for non-obvious systems (floating block, U-tube, magnet, charged bead); effective spring constant of series/parallel/combined springs; energy-partition and phase questions; loss-of-contact and maximum-amplitude conditions; and superposition/Lissajous problems.

Exam tricks & shortcuts

• Reduce any oscillator to a = −ω²x; then T = 2π/ω instantly.
• Springs: series adds compliances (1/k), parallel adds stiffnesses (k).
• v_max = ωA, a_max = ω²A — quick sanity checks.
• Mnemonic: "Restoring and proportional ⇒ SHM."

Key formulas

• SHM: a = −ω²x; x = A cos(ωt + φ); v = ±ω√(A²−x²); v_max = Aω.
• Energy: total E = ½mω²A² = ½kA²; KE = ½mω²(A²−x²), PE = ½mω²x².
• Spring: T = 2π√(m/k); pendulum: T = 2π√(L/g).
• Springs: series 1/k = Σ1/kᵢ; parallel k = Σkᵢ.
• Damped: A = A₀e^(−bt/2m); resonance at driving = natural frequency.

SHM has a restoring force ∝ displacement; energy shuttles between kinetic and potential while the total stays constant.

Worked Example

Problem: A 1 kg block is connected to two springs of stiffness k₁ = 300 N/m and k₂ = 600 N/m. Find the period of oscillation when the springs are connected (a) in series and (b) in parallel with the block.

Solution:
The period of a mass–spring oscillator is T = 2π√(m/k_eff), so find the effective stiffness in each case.

(a) Springs in series share the same force but stretch additively, so:
1/k_eff = 1/k₁ + 1/k₂ = 1/300 + 1/600 = 2/600 + 1/600 = 3/600
k_eff = 200 N/m.
T = 2π√(1/200) = 2π × 0.0707 ≈ 0.444 s.

(b) Springs in parallel stretch by the same amount, so their stiffnesses add:
k_eff = k₁ + k₂ = 300 + 600 = 900 N/m.
T = 2π√(1/900) = 2π/30 ≈ 0.209 s.

The parallel arrangement is stiffer, giving a shorter period.

Answer: (a) T ≈ 0.444 s (series); (b) T ≈ 0.209 s (parallel).