Mechanical oscillations

At an endpoint, a mass on a spring stops momentarily, yet its velocity changes fastest there. At the centre, the situation is reversed. We will connect these changing roles to phase, force and energy exchange.

Release a displaced mass: the spring pulls it towards equilibrium, and inertia carries it through to the other side. Its states repeat: this is oscillation. A complete cycle restores both position and direction of motion. If timing starts as it passes the centre to the right, the next leftward crossing completes only half a cycle. A single photograph of position cannot identify its stage in the repetition.

Time n complete cycles over t seconds. One cycle takes T=t/n and the number per second is \nu=n/t=1/T. T is in seconds and ν in hertz: 1 Гц=1 с⁻¹. Doubling frequency halves the period. Timing several cycles reduces the relative effect of starting the stopwatch inaccurately, but cannot correct a mistaken cycle count. Here n counts cycles, not the turns of the previous lesson.

Set equilibrium at x=0. The sign of displacement x indicates which side the mass is on. Amplitude xₘ is the greatest displacement magnitude, half the distance between endpoints. Both are measured in metres. Doubling amplitude doubles the full excursion. Negative x does not mean negative amplitude, and zero average position does not mean no motion.

Phase specifies the current stage of a cycle. A full repeat corresponds to 2π radians; initial phase φ₀ sets the state at t=0. For uniformly advancing phase, its rate is \omega=2\pi\nu=2\pi/T in radians per second. Doubling ω completes a cycle in half the time. Frequency in hertz and angular frequency describe the same rhythm, but their numerical values differ by a factor of 2π.

An ideal linear restoring force gives x=x_m\sin(\omega t+\varphi_0), harmonic motion with constant amplitude and frequency. Sine lies between −1 and +1, so x stays between −xₘ and +xₘ. Changing xₘ scales the graph vertically; ω changes its time scale, and φ₀ shifts the starting phase. The argument is in radians. At φ₀=0, the mass starts by crossing equilibrium in the positive direction.

The spring pulls harder with greater displacement, while mass resists velocity change. For an ideal massless Hooke’s-law spring without losses, T=2\pi\sqrt{m/k}, with m in kilograms and k in newtons per metre. Four times the mass doubles the period; four times the stiffness halves it. Amplitude does not enter as long as the force is linear. Large deformations can invalidate this independence.

Gravity provides a simple pendulum’s restoring action. For a point particle on a light inextensible string at small angles, T=2\pi\sqrt{l/g}. Length l in metres runs from the pivot to the mass centre; g is in m/s². A longer pendulum oscillates more slowly; larger g speeds its return. Mass cancels because weight and inertia increase together. Large angles and air resistance fall outside this formula’s conditions.

Why is speed greatest at the centre? The restoring force has accelerated the mass up to this point and has not yet begun slowing its motion towards the other end. For harmonic motion, velocity amplitude is v_m=x_m\omega in metres per second. At the same amplitude, doubling frequency requires twice the speed, so vₘ doubles. At the endpoints speed is zero, but the mass does not stay there: force starts its return.

In an ideal oscillator force opposes displacement, so a=-\omega^2x and its greatest magnitude is a_m=x_m\omega^2. Doubling ω at the same amplitude quadruples acceleration magnitude; its unit is m/s². At the endpoints |x| and |a| are greatest but velocity is zero. At equilibrium a=0 and speed is greatest. Zero velocity and zero acceleration describe different states; one cannot be inferred from the other.

In an ideal spring system total energy is constant: E_0=mv_m^2/2=kx_m^2/2 in joules. At an endpoint it is entirely potential; at equilibrium entirely kinetic. The equality compares maxima reached at different times, not a sum of simultaneous maxima. Doubling amplitude quadruples the total energy. Without losses, the same energy store repeatedly changes form.

Take a snapshot between an endpoint and equilibrium: the spring is still deformed and the mass is already moving. Now E_0=mv^2/2+kx^2/2, with v and x at the same instant. This sum stays constant in the ideal model. Friction converts some mechanical energy into internal energy, reducing the mechanical store and the amplitude. Adding the two maximum energies would wrongly double the initial store.

A mass on an ideal horizontal spring of stiffness 100 N/m oscillates with amplitude 0.10 m. Find total energy and kinetic energy at equilibrium.

Start at an endpoint: speed is zero, so total energy equals the spring’s energy at xₘ = 0,10 м.

E₀ = kxₘ²/2 = 100 × 0,10²/2 = 0,50 Дж.

At equilibrium x = 0, the horizontal spring is undeformed and has zero potential energy. Therefore Eк = E₀ = 0,50 Дж.

Do not add these values: they describe the same energy store in different forms at different times.

Total energy: E = 0,50 Дж. At equilibrium it is all kinetic: Eк = 0,50 Дж.

An oscillation repeats the full state. Passing the same position in the opposite direction is not a complete cycle.

Phase, amplitude and frequency describe different aspects of harmonic motion.

Speed is greatest at the centre, acceleration magnitude at the endpoints. A fixed energy store moves between motion and deformation.

Next we find a similar energy exchange in an electric circuit.