Oscillations
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Direct answer
Pull a mass on a spring to one side and release it: the acceleration always points back toward the mean position with a = -ω^2x, and that is simple harmonic motion — sinusoidal, isochronous (period independent of amplitude), and the model behind pendulums, vibrating strings and half of acoustics. The two systems NEET-UG actually examines are the spring-mass oscillator, T = 2π sqrt(m/k), and the small-angle pendulum, T = 2π sqrt(l/g), followed by damped motion and resonance.
What you must remember
- x(t) = A sin(ωt + φ): speed is maximum (Aω) at the mean position where displacement and acceleration are zero; acceleration is maximum (Aω^2) at the extremes where speed is zero.
- Spring-mass period T = 2π sqrt(m/k): heavier oscillates slower, stiffer springs faster; amplitude has no say (for ideal SHM).
- Simple pendulum T = 2π sqrt(l/g): independent of the bob's mass; the seconds pendulum (T = 2 s) needs a length of about one metre.
- Total energy E = (1/2)kA^2 = (1/2)mω^2A^2, sloshing between kinetic (max at the centre) and potential (max at the extremes); the two are equal at x = ±A/sqrt(2).
- Timing facts: mean position to an extreme takes T/4; extreme to extreme T/2; a full out-and-back T.
- Damped oscillations decay in amplitude (energy leaks to friction); critical damping returns the system to rest fastest without oscillating — the dead-beat galvanometer design.
- Forced oscillations respond to a driving frequency; resonance — driving frequency equal to natural frequency — makes the amplitude peak sharply when damping is small, which is why troops break step over bridges.
- SHM is the projection of uniform circular motion, so ω in SHM carries the same meaning as angular speed in that circle.
Worked example: energy audit of an oscillator
A 2 kg block on a spring of stiffness 200 N/m is pulled 10 cm from equilibrium and released. The angular frequency is ω = sqrt(k/m) = sqrt(200/2) = 10 rad/s, so the period is T = 2π/10 ≈ 0.63 s — note how neither the 10 cm nor any later amplitude appears in T. The total energy is E = (1/2)kA^2 = (1/2) × 200 × 0.01 = 1 J. At the halfway displacement x = 5 cm, the stored potential energy is (1/2) × 200 × (0.05)^2 = 0.25 J, so the kinetic energy there must be 0.75 J and the speed v = sqrt(2 × 0.75/2) ≈ 0.87 m/s. Cross-check with the kinematic formula v = ω sqrt(A^2 - x^2) = 10 × sqrt(0.01 - 0.0025) = 10 × 0.0866 = 0.87 m/s: the energy ledger and the motion formula agree exactly, and at the 0.707A displacement the split would be 50-50. Both routes to the same number is the habit that converts SHM questions into thirty-second answers.
Where students slip
"Heavier pendulum swings slower" — false; mass cancels in T = 2π sqrt(l/g), because the driving weight and the inertia it must move are the same m. The period-independence of amplitude holds only for small angles, where sin θ ≈ θ; swing the pendulum wide and the period lengthens, which is why the practical insists on amplitudes of a few degrees. In experiments, timing one oscillation multiplies reaction-time error by a factor of 20 relative to timing twenty and dividing — the exam states this as a percentage-error question. The damped case confuses students about period: weak damping barely shifts the frequency while the amplitude decays exponentially, so "damping changes the period noticeably" is false as stated. Finally, do not exchange ω and f: the mains of acoustics questions quote frequency in hertz, and ω = 2πf is the conversion SHM formulas silently assume.
Frequently asked questions
At what displacement are kinetic and potential energies equal in SHM?
At x = A/sqrt(2) ≈ 0.707A, where each is half of the total (1/2)kA^2; at the centre all energy is kinetic, at the extremes all potential.
Does a pendulum's period depend on the mass of the bob?
No: T = 2π sqrt(l/g). Gravity pulls harder on a heavier bob, but the same extra mass resists acceleration, and the effects cancel.
What is resonance?
The driving frequency coincides with the system's natural frequency, so energy accumulates and the steady-state amplitude becomes very large when damping is small — sharpness of resonance falls as damping rises.
Why does a pendulum clock taken from the hills to the equator run slow?
g is smaller at the equator, so T = 2π sqrt(l/g) lengthens and each swing takes longer — the clock completes fewer swings per day and loses time.
What is critical damping?
The damping for which the system returns to equilibrium in the shortest time without oscillating — the design goal of vehicle shock absorbers and dead-beat galvanometers.