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SHM

Part of the NEET UG study roadmap. Physics topic phy-013 of Physics.

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SHM

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your NEET UG Physics paper.

Simple Harmonic Motion (SHM) is periodic motion where the restoring force is directly proportional to the displacement from the mean position and opposite in direction: F = −kx.

The displacement follows a sinusoid: x(t) = A sin(ωt + φ), where A is amplitude, ω is angular frequency, and φ is the initial phase. Velocity and acceleration are sinusoidal too, separated by π/2 and π in phase respectively.

Must-remember formulas for NEET:

  • Spring-mass: T = 2π√(m/k)
  • Simple pendulum: T = 2π√(L/g) (small-angle approximation only)
  • Total energy: E = ½kA²
  • v_max = ωA, a_max = ω²A

High-yield tip: A body is at rest only at the extremes (x = ±A) where acceleration is maximum; v_max occurs at the mean position.

Springs in series give 1/k = 1/k₁ + 1/k₂; in parallel, k = k₁ + k₂. SHM also describes the LC circuit, which is why this chapter directly feeds into AC and wave problems.

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months before NEET UG.

Defining condition and reference circle

SHM is defined by the single force law F = −kx. Any system obeying this linear restoring force executes SHM. Geometrically, SHM is the projection of uniform circular motion onto a diameter: a particle moving in a circle of radius A at constant angular speed ω has its x-coordinate trace out A sin(ωt + φ).

Phase relationships

Starting from x = A sin(ωt + φ), differentiating once and twice:

  • v = dx/dt = Aω cos(ωt + φ)
  • a = dv/dt = −Aω² sin(ωt + φ) = −ω²x

So v leads x by π/2, and a is π out of phase with x (a opposes displacement). This phase picture is the basis of most energy and graph-based NEET MCQs.

Time periods and energy

For a spring-mass system, substituting k = mω² into ω = √(k/m) gives T = 2π√(m/k). For a simple pendulum valid only when the angular swing θ ≲ 15°, T = 2π√(L/g). Total mechanical energy is constant: E = ½kA² = ½mω²A², continuously splitting between KE and PE.

QuantityMean position (x = 0)Extreme position (x = ±A)
Displacement0±A
Velocityv_max = ωA0
Acceleration0a_max = ω²A
Kinetic energyMaximum (E)0
Potential energy0Maximum (E)

Typical NEET question patterns

Expect one MCQ testing: (1) the time-period formula for spring or pendulum, (2) phase-difference identification from graphs, (3) spring combinations (series/parallel), or (4) energy distribution between KE and PE at a given displacement.

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Generalised SHM and effective spring constant

Beyond a single spring, SHM arises in any system where the restoring “force-like” variable is linearly opposed to the displacement-like variable. This includes the torsional pendulum (τ = −Cθ, giving T = 2π√(I/C)), a liquid column in a U-tube oscillating vertically, and the LC circuit where charge q plays the role of x with effective k = 1/C and effective m = L, yielding T = 2π√(LC).

Spring combinations modifying T

When two springs of constants k₁ and k₂ are attached to the same mass:

  • Parallel (both stretched equally): k_eff = k₁ + k₂, so T decreases by a factor 1/√(1 + k₂/k₁).
  • Series (same force, additive extension): 1/k_eff = 1/k₁ + 1/k₂, so T increases.

A half-cut spring (each half has constant 2k) halves the period when used in place of the original.

Common pitfalls

  • Writing F = +kx drops the directional sign — F must point toward equilibrium, hence the negative.
  • The simple pendulum formula is valid only for small angles; at θ = 60° the true period is roughly 7% longer.
  • Confusing angular frequency ω (rad/s) with ordinary frequency f (Hz): remember ω = 2πf.
  • For a physical (compound) pendulum, use distance from pivot to centre of mass as effective L, not the rod’s total length.

Worked micro-example

A 0.5 kg mass on a spring of k = 200 N/m is displaced 3 cm and released. Then ω = √(200/0.5) = 20 rad/s, T = 2π/20 ≈ 0.314 s, v_max = 20 × 0.03 = 0.6 m/s, E = ½ × 200 × (0.03)² = 0.09 J. At x = 2 cm, PE = ½ × 200 × (0.02)² = 0.04 J, KE = 0.09 − 0.04 = 0.05 J, v = √(2 × 0.05/0.5) = 0.447 m/s.

Practice prompts

  1. Two identical springs (k each) are joined end-to-end and attached to a mass m. Find the new period in terms of the original T₀.
  2. A simple pendulum of length 1 m has a bob of mass 0.2 kg displaced to 10° and released. Compute its speed and acceleration at the lowest point, and verify T using the small-angle formula.

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