Oscillations

01

Describe and measure an oscillation

An oscillation repeats about an equilibrium position. To describe it, identify that position, measure displacement from it and count a complete cycle rather than every crossing of the centre.

Equilibrium and free oscillation

At the equilibrium position, the body can remain at rest with forces balanced when no periodic drive is applied. Displacement x is the signed distance from this position. A body can pass through equilibrium at nonzero speed; being at that position does not mean it is at rest.

A horizontal cart attached to a spring can oscillate after a displacement and release. In the ideal free model, no energy is gained from or lost to the environment: kinetic and potential energy exchange while their total remains constant. Restoring forces still act; there is no continuing periodic drive.

A small-angle pendulum is another approximate example when losses are negligible. Real carts and pendulums gradually lose mechanical energy unless supplied with more. Their damped responses are considered later in the chapter.

Amplitude, period and the two frequency quantities

The amplitude x₀ is the maximum magnitude of displacement, in metres. It is non-negative. If a rest pointer is at 20.0 cm and the extremes are 15.0 cm and 25.0 cm, the amplitude is 5.0 cm = 0.050 m, not the 10.0 cm distance between extremes.

The period T is the time for one complete cycle, in seconds. The system must return to the same state of motion: two centre crossings in the same direction are one period apart. Consecutive crossings in opposite directions are half a period apart.

The frequency f is cycles per second, in hertz (Hz = s^-1). The angular frequency ω describes the rate at which the sinusoidal phase advances, in rad/s. One cycle corresponds to 2π radians:

Frequency and angular frequency have different numerical values and units. If ω is in rad/s and t is in seconds, the phase ωt is in radians.

Estimate the scale before calculating

A few-centimetre amplitude and roughly 20 cycles in 30 s suggest a period of order 1 s, frequency of order 1 Hz and angular frequency of a few rad/s. These are rough scales, not a reason to make f and ω numerically equal.

Measure displacement and timing together

1. Establish equilibrium. Let the undriven system settle and mark the rest reading. Calibrate a displacement scale in the plane of motion. Subtract the equilibrium reading from position to obtain signed x.
2. Use a reproducible release. Keep the spring attachments and supports secure, remain within the apparatus range and release without a continuing push.
3. Time a stated number of full cycles. Use crossings of the same reference in the same direction, then repeat. Choose a stopwatch range and resolution suitable for the full timing interval.
4. Record position against time. A suitable position sensor or calibrated video can give an x-t record. Track one point, retain the sample or frame times, view perpendicular to the motion plane and calibrate distance in that plane.
5. Check the model conditions. Compare successive cycles for changing period or amplitude. Keep mass and restoring arrangement fixed when comparing amplitudes, and remain within the linear or small-angle range.

Many-cycle timing reduces the fractional effect of start/stop timing. Repeats reveal variation but do not correct a miscount, the wrong equilibrium reference or perspective error. A short, nearly sinusoidal record does not establish constant amplitude over a much longer interval.

The motion-graph method shows how calibrated positions can also give approximate velocities and accelerations.

Phase specifies the stage of the cycle

Phase expresses the stage of an oscillation as an angle relative to a chosen reference. One cycle advances phase by 2π rad. Phase difference compares two stages; for traces at the same frequency, a positive time separation Δt between corresponding stages gives the phase-difference magnitude:

State which trace reaches a corresponding stage first. A quarter-cycle advance is π/2 rad; half a cycle is π rad. Equal displacement at one instant does not establish equal phase, because the bodies may be moving in opposite directions.

For x₁ = x₀ sin(ωt) and x₂ = x₀ sin(ωt + π/2), x₂ leads x₁ by T/4. At t = 0, x₂ is at its positive extreme while x₁ crosses equilibrium towards positive x.

02

Recognise the SHM condition

Simple harmonic motion requires acceleration proportional to displacement from equilibrium and directed towards equilibrium. A restoring direction alone is not enough.

Let x be signed displacement from equilibrium, positive to the right. The defining equation is:

The constant ω is angular frequency, in rad/s. When x is positive, a is negative; when x is negative, a is positive. At x = 0, the restoring acceleration is zero. The minus sign makes the acceleration restoring, while the factor ω² gives its proportionality to x.

An a-x graph for SHM is a straight line through the origin with negative gradient -ω², in s^-2. A positive gradient would accelerate away from equilibrium. A curved restoring relationship is not this linear model.

If measured position was not referenced to equilibrium, the a-position graph may have a shifted intercept. Correct the displacement reference rather than assuming a nonzero intercept is the SHM equilibrium.

Choose the time origin before using a sine model

Use amplitude x₀ = 0.0500 m. Choose t = 0 when the cart crosses equilibrium moving to the right. A suitable displacement model is:

The phase 4.00t is in radians. At t = 0, x = 0 and the displacement graph is rising. The symbol x₀ is amplitude, not the initial displacement in this convention.

If instead t = 0 were chosen at release from the positive extreme at rest, x = x₀ cos(ωt) would describe the same ideal oscillator with a different phase. The clock zero does not have to be the instant the practical apparatus was released; a later rightward centre crossing can define it.

For the sine choice, the displacement gradient is greatest and positive at the first centre crossing, zero at the positive extreme, and negative during the return. Its velocity and acceleration traces satisfy a = -ω²x throughout, as the aligned graphs show.

03

Connect velocity and acceleration

Velocity is the gradient of displacement against time, and acceleration is the gradient of velocity against time. Use those relationships to check both signs and the timing of maxima or zeros.

For ideal SHM with x = x₀ sin(ωt), choose t = 0 at a rightward equilibrium crossing. Then:

Here v₀ is the velocity amplitude, or maximum speed. It is also the initial velocity only for this chosen phase. Displacement, velocity and acceleration are signed; their amplitudes are non-negative.

Read three aligned graphs

Use x₀ = 0.0500 m, ω = 4.00 rad/s and T = 1.57080 s. The model is:

At t = 0 the cart crosses the centre to the right. At T/4 it turns at the positive extreme; at T/2 it crosses the centre to the left. At 3T/4 it turns at the negative extreme, and at T the original state repeats.

At the centre: speed is greatest and restoring acceleration is zero. At either extreme: velocity is momentarily zero, but acceleration has greatest magnitude towards equilibrium. Zero instantaneous velocity is therefore not enough to identify equilibrium.

Comparing corresponding stages of the sinusoidal traces, velocity leads displacement by π/2 rad. Acceleration and displacement are in antiphase, separated by π rad: when x is most positive, a is most negative. Their different units do not share one common vertical scale.

When x is positive, acceleration is negative on both passes. It slows a cart moving right, but speeds a cart returning left. Acceleration is set by displacement, not by a rule that it must always oppose velocity.

Find velocity from position and direction

Using sin²(ωt) + cos²(ωt) = 1 to eliminate time gives:

The square root gives speed magnitude. Select + for motion towards positive x and - for motion towards negative x. The model requires |x| ≤ x₀; a position beyond the stated amplitude does not belong to this motion.

Estimate rates from a position record

The following five positions are generated from the same sine model and rounded to 0.00001 m. They illustrate the calculation method; they are not observations that independently test the model.

For each adjacent interval, calculate Δx/Δt. The four average velocities are 0.1947, 0.1640, 0.1073 and 0.0338 m/s, assigned approximately to interval midpoints 0.05, 0.15, 0.25 and 0.35 s. Keep these derived values separate from the original positions.

For example, the velocity over 0.10 to 0.20 s is (0.03587 - 0.01947)/0.10 = 0.1640 m/s. The next interval gives (0.04660 - 0.03587)/0.10 = 0.1073 m/s.

Pair this estimate with x = 0.03587 m at 0.20 s. The exact sine model gives about -0.573885 m/s² there. Finite intervals and position rounding explain why the estimate differs; it is not the exact instantaneous acceleration.

To investigate real motion, apply the method to an actual calibrated x-t record spanning both sides of equilibrium. Compare the observed displacement pattern with a sinusoid, then plot acceleration estimates against displacements at corresponding times. A roughly straight a-x relationship through the origin with negative gradient supports the SHM model over the tested range. Its gradient estimates -ω², which can be compared with the independently timed period.

The few positive-side estimates in this generated illustration do not establish that relationship experimentally and do not need a fitted line. For real records, check calibration, equilibrium zero, frame timing, changing amplitude and possible nonlinearity.

Shorter intervals can improve a finite-gradient approximation, but differences then become smaller relative to measurement resolution. Estimating rates amplifies noise, especially when velocity estimates are differenced again. Choose intervals that resolve the motion while still giving reliable position and timing differences; smaller is not automatically better.

04

Follow the energy exchange

An ideal free oscillator repeatedly exchanges kinetic and potential energy. Their total stays constant when no energy is supplied or dissipated.

Choose the horizontal spring and cart as the system. Take elastic potential energy at equilibrium as zero. The cart stays at one height, so gravitational energy is unchanged. Use a linear spring, negligible losses and no continuing drive.

Obtain the potential energy from work

The work needed to stretch or compress a linear spring from equilibrium is the area under its force-extension graph. This gives:

The second form uses k = mω² for this spring-cart model. Squaring displacement makes E_p non-negative on either side of equilibrium. At either extreme, velocity is zero, so the total energy is:

This is derived from the spring work and the turning-point condition. It is not an additional energy to add to E_k + E_p.

Over the allowed displacement range, E_p is an upward parabola and E_k is the total minus that parabola. Both remain non-negative. The energy pattern repeats after T/2 because the squares x² and v² repeat then, even though the full signed motion takes T.

Change amplitude within the same linear model

Doubling amplitude to 0.100 m with the same mass and stiffness doubles maximum speed from 0.200 to 0.400 m/s and maximum acceleration magnitude from 0.800 to 1.60 m/s².

The period stays the same only while the same linear restoring model remains valid. A larger displacement that changes the restoring relationship need not preserve that result.

For a vertical oscillator, the potential-energy account must include both spring and gravitational contributions about the shifted equilibrium. Do not label all of that combined change as spring energy alone.

05

Compare damped responses

Damping transfers mechanical energy from the oscillator to other forms, often internal energy in the apparatus and surroundings. The response depends on the degree of damping.

The ideal free spring-cart model has constant amplitude and no energy loss. With damping, a resisting force acts as well as the restoring force. The ideal acceleration relationship a = -ω²x alone no longer describes the whole resultant.

In an undriven oscillating response, decreasing displacement peaks show decreasing mechanical energy. Energy is transferred, not destroyed. Passing through zero displacement is not evidence of zero mechanical energy: the body can still be moving.

Compare the same release conditions

The three model responses below use the same mass and stiffness. Each starts at +0.0500 m displacement with zero velocity, then receives no continuing drive. This time origin is release from an extreme, unlike the earlier centre-crossing sine model.

• Light or underdamping: the body crosses equilibrium repeatedly, with decreasing oscillation amplitude.
• Critical damping: the boundary case gives the most rapid return towards equilibrium without oscillating for this release-from-rest comparison.
• Heavy or overdamping: the return is nonoscillatory but slower under the same initial conditions. More damping does not always mean a quicker return.

The nonoscillating traces approach equilibrium; the model does not assign an exact finite stopping time merely because the curve becomes close to zero. The comparison is tied to the stated initial displacement and zero initial velocity, not a claim that critical or overdamped systems can never cross equilibrium under any possible initial motion.

Apply the comparison to a suspension

A car's springs allow movement after a disturbance such as a road bump. Dampers transfer mechanical energy to internal energy and limit continued bouncing.

Too little damping allows repeated motion. Excessive damping can make the return slow. A near-critical model illustrates the aim of rapid settling without continued oscillation. A real suspension has several moving parts, so this single-response comparison does not imply that every part of it is exactly critically damped.

A measuring-instrument pointer provides another application: damping can give a rapid, readable approach to the final position instead of repeated overshoot. The desired response involves both avoiding oscillation and reaching a useful reading promptly.

06

Driving and resonance

A periodic external force can keep an oscillator moving by supplying energy. The size of the response depends on driving frequency, damping and drive strength.

Separate the driving frequency from the natural frequency

The natural frequency is the frequency of free oscillation. For the undamped spring-cart model with m = 0.400 kg and k = 6.40 N/m, ω₀ = 4.00 rad/s and f₀ = ω₀/(2π) = 0.636620 Hz.

The driving frequency is set by the external periodic force. After the initial transient has died away, the steady forced response of the linear oscillator repeats at that driving frequency. Natural and driving frequency are different concepts even when their numerical values coincide.

With damping present, maintaining a steady amplitude requires energy input. Over each steady cycle, energy supplied equals energy dissipated. Constant amplitude does not mean that damping has stopped acting.

Read an amplitude-frequency response

Vary the driving frequency while keeping the driving-force amplitude and the oscillator fixed. At resonance, the response amplitude reaches a maximum when the driving frequency is close to or at the natural frequency.

The vertical axis below is response amplitude, not signed displacement. The horizontal axis is driving frequency in Hz, not time. Each point represents the steady response at one chosen driving frequency.

These supplied model curves use m = 0.400 kg, k = 6.40 N/m and sinusoidal driving-force amplitude 0.0128 N. They are predictions for a controlled comparison, not measured responses.

Curve A, with the least damping of the three, has the highest, sharpest peak: about 12.5 mm near 0.633 Hz. The intermediate curve B peaks at about 5.65 mm near 0.616 Hz. Curve C, with the greatest damping, has a lower, broader peak: about 3.49 mm near 0.576 Hz.

Greater damping reduces the large response and makes the resonance less sharp. In this displacement-response comparison, it also moves the peak farther below the undamped f₀. Do not force every damped curve to peak at exactly that reference frequency.

At very low driving frequency, the displacement follows the slowly varying force almost statically. The limiting amplitude here is F₀/k = 0.0128/6.40 = 0.00200 m = 2.00 mm, so the response curves do not begin with a fabricated zero-amplitude limit.

The driven amplitude is determined by these drive and damping conditions; it need not equal the 0.0500 m amplitude chosen for the earlier free motion. In a perfectly lossless oscillator driven at exact resonance, a finite steady amplitude would not be established: continued input increases the response. The finite peaks above include damping.

Useful and unwanted resonance

A swing: the swing is the responding system and the repeated pushes are the drive. Well-timed pushes do positive work each cycle and build its motion when their timing is close to the free-oscillation rhythm. A maintained amplitude requires the pushes to replace losses.

A musical resonator: a driving vibration can produce a strong response near one of the resonator's natural frequencies. Energy passes into its vibration and then into the surrounding air, supporting sound production.

A machine support: a running motor can exert a periodic force. If its driving frequency is near a support's natural frequency, a large response can cause unwanted vibration and stress. Damping can reduce the response, while a suitable design or operating change can separate the driving and natural frequencies.

Changing mass or stiffness changes the system's natural frequency, but either change is not automatically an improvement. Judge whether it moves the system towards or away from the actual drive frequency.

Investigate the response with controlled conditions

Change the driver frequency in steps, let transient motion settle, and record the steady amplitude from the equilibrium position. Use the same calibrated displacement method and measure enough of the motion to distinguish the steady amplitude from a brief transient peak.

Keep drive strength, mass, restoring arrangement and damping fixed when obtaining one frequency-response curve. To compare damping, change damping deliberately while maintaining the other conditions, then repeat the frequency scan.

Record what the apparatus controls. A driver that maintains a support's displacement amplitude does not automatically maintain driving-force amplitude, so its curve should not be claimed to reproduce this fixed-force comparison without checking that condition.