Spring Oscillations
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Direct answer
Hang a mass from a spring, pull it down and let go: it bounces with period T = 2π√(m/k), where k is the spring constant in N/m — the restoring law F = −kx makes the motion simple harmonic, with ω = √(k/m). Mass and stiffness share the honours: quadruple the mass and the period doubles; quadruple k and it halves. Gravity shifts the equilibrium position but never the period, because it merely relocates the point where x = 0 is measured. Combinations follow capacitor logic: parallel springs add, k = k1 + k2 (stiffer, faster), series springs add reciprocals, 1/k = 1/k1 + 1/k2 (softer, slower) — the most tested add-on in this block.
What you must remember
- Core equations: F = −kx; T = 2π√(m/k); ω = √(k/m); acceleration a = −(k/m)x; frequency = (1/2π)√(k/m).
- Gravity neutrality: in vertical springs mg only displaces the equilibrium by mg/k; the period is untouched — the classic assertion-reason line.
- Parallel combination: k_eff = k1 + k2, both springs stretching the same amount and sharing the load.
- Series combination: 1/k_eff = 1/k1 + 1/k2, the same force extending each spring by its own amount.
- Cut-spring result: halving a spring's length doubles its constant (k ∝ 1/L for the same wire); the same mass then oscillates with T/√2.
- Energy pairing: U = ½kx²; total E = ½kA²; speed at displacement x is v = ω√(A² − x²), maximum v = Aω at equilibrium.
- Stiffness reading: k = force per unit extension; a spring pulling 100 N per centimetre of stretch has k = 10^4 N/m.
- Contrast with the pendulum: spring period is independent of g (works in zero gravity), the pendulum's is set by g — a comparison NEET enjoys.
Working the numbers
A 0.5 kg block on a frictionless table against a spring of k = 200 N/m: T = 2π√(0.5/200) = 2π × 0.05 ≈ 0.31 s, a frequency of about 3.2 Hz. Pull it 5 cm and release: stored energy E = ½ × 200 × 0.05² = 0.25 J, all kinetic at the centre, so v_max = Aω = 0.05 × √(200/0.5) = 0.05 × 20 = 1 m/s — the energy check ½mv² = 0.25 J confirms it. Now replace the spring by two springs, 200 and 300 N/m. In series: 1/k = 1/200 + 1/300 gives k_eff = 120 N/m and the period stretches by √(200/120) ≈ 1.29 times; in parallel k_eff = 500 N/m and the period shrinks. Same arithmetic, opposite verdicts — exactly the discriminations the options are built from.
Where NEET sets the trap
Series and parallel are regularly swapped, because resistor intuition fails here — springs combine like capacitors, and candidates who memorised resistor rules choose the wrong k_eff every time. The gravity assertion returns yearly: hang the same spring-mass system vertically and the period does not change; only the resting position shifts by mg/k. The cut-spring question is a two-step trap: cutting in half gives k′ = 2k, then the new period is T/√2, and options include T/2 (wrong power) and 2T (wrong direction). Unit slips with centimetres against N/m cost a factor of 10² silently. A final comparative favourite: the same apparatus carried to the Moon keeps identical time (no g anywhere in T), while the pendulum slows — one question, two chapters, four options.
Frequently asked questions
What is the period of a 0.5 kg mass on a 200 N/m spring?
T = 2π√(m/k) = 2π√(0.5/200) ≈ 0.31 s, a frequency of about 3.2 Hz.
How do spring constants combine in series and in parallel?
Parallel: k = k1 + k2 (stiffer); series: 1/k = 1/k1 + 1/k2 (softer) — the same rules as capacitors, the reverse of resistor intuition.
Does gravity change the period of a vertical spring oscillator?
No; gravity only shifts the equilibrium by mg/k, and the period stays 2π√(m/k) about the new centre.
A spring is cut into two equal halves. What is each half's constant?
Twice the original, since k ∝ 1/L for the same wire; the same mass on one half oscillates with period T/√2.
Where is the block's speed maximum in spring SHM?
At the equilibrium position, with v_max = Aω = A√(k/m); at the two extremes the block momentarily stops.