Oscillation parameters and pendulums

A swing tends towards its lowest point but does not stop when it reaches it. Why does a restoring effect not bring immediate rest, and what determines how long the motion takes to repeat?

Pull a mass attached to a spring aside and release it. The restoring force accelerates it towards stable equilibrium. At equilibrium the mass is already moving: the restoring force vanishing at that instant does not remove its velocity. Inertia carries it past equilibrium, and the force now points back. Together, restoring action and inertia produce oscillations. Without inertia it would not pass equilibrium; without restoring action it would not turn back.

Follow the oscillator's energy. At an extreme, speed is momentarily zero and potential energy is greatest. Towards equilibrium, potential energy becomes kinetic energy; speed is greatest there. The exchange then reverses. With small losses the cycle nearly repeats. Energy is not created on each pass: the system uses energy supplied by the initial displacement, and losses gradually reduce this store.

How do we compare how far two masses swing? Amplitude A is the greatest displacement magnitude from equilibrium, measured in metres in SI. For extremes −3 cm and +3 cm, A = 3 см, while the full range is 6 cm. Doubling amplitude doubles that range but does not by itself specify cycle time. The average coordinate can be zero even for large oscillations.

A mass passes equilibrium twice each cycle, once rightwards and once leftwards. Is each pass a full oscillation? No: a full cycle returns to the same position with the same direction of motion. Period T is this cycle time in seconds. Going from the left extreme to the right takes only T/2; the return is still needed. Starting the count at an extreme makes the direction change easy to see.

For a short period, human reaction time when starting a stopwatch may be large compared with one cycle. Measure total time t for N full oscillations and divide: T = t/N. Here N is a dimensionless cycle count, not the power symbol from the previous lesson. Twice as many cycles take twice as long while period stays the same. This reduces the relative starting error, but cannot correct miscounting or systematic clock error.

The same rhythm can be described by the number of full cycles per second. Frequency is ν = N/t = 1/T, measured in hertz: 1 Гц = 1 с⁻¹. At T = 0,5 с, ν = 2 Гц. A longer cycle means fewer cycles per second: doubling T halves ν. Frequency and period are not independent; they describe the same repetition rate in two ways.

To calculate a string pendulum's period, idealise it as a small heavy bob on a massless, inextensible string with little resistance. This is a simple pendulum. Length l runs from the suspension point to the bob's centre. At small angles the restoring effect is approximately proportional to displacement, giving T = 2π√(l/g). The ratio under the root has units s². The model explains the rhythm but need not accurately describe wide swings of a real swing.

Replace the bob with a heavier one at the same length. Gravity and inertia increase together, so mass cancels from the period. A string four times as long gives twice the period because of the square root. Greater g strengthens the return and shortens the period. Amplitude is absent only in the small-angle approximation: greatly widening the swing violates this condition and changes the period.

For a mass on an ideal spring, spring constant k sets the restoring effect while mass m resists changes in velocity. Their balance gives T = 2π√(m/k), with m in kilograms and k in N/m. Quadrupling mass doubles the period; quadrupling the spring constant halves it. Hooke's law applies, so this model's period is independent of amplitude while deformation remains within the linear range.

Turn a horizontal mass–spring oscillator upright. The mass stretches the spring to a new equilibrium where elasticity balances gravity. For a small displacement from there, the additional restoring force is again determined by k, so the period remains 2π√(m/k). The model assumes negligible spring mass and losses. A heavy spring, friction or departure from Hooke's law requires a refined calculation.

A mass completes 12 full oscillations in 18 s. Find its period and frequency.

Distinguish the time for one cycle from cycles per second. We measured 12 full cycles in 18 s.

T = t/N = 18/12 = 1,5 с: the average time to return to the same state.

ν = N/t = 12/18 ≈ 0,67 Гц. The reciprocal check gives 1/T = 1/1,5 ≈ 0,67 Гц.

The period exceeds one second, so less than one full cycle occurs per second; νT equals one within rounding accuracy.

T = 1,5 с, ν ≈ 0,67 Гц. One cycle takes one and a half seconds; about two thirds of a cycle occur per second.

Oscillations arise from restoring action together with inertia; energy changes between mechanical forms.

Amplitude describes displacement size; period and frequency describe repetition, with ν = 1/T.

Simple-pendulum and mass–spring periods belong to different models, each with explicit limitations.

Next we distinguish free, damped and forced oscillations.