Simple Harmonic Motion and Oscillations

Oscillation means repeated motion around an equilibrium position. Periodic motion repeats its state after a fixed time interval.

A rotating wheel can be periodic without moving back and forth along one line. A damped swing can oscillate without repeating exactly the same amplitude.

For one-dimensional SHM, measure displacement x from equilibrium. Then the defining acceleration law is:

a = −ω²x

Here, a is acceleration and ω is angular frequency. The coefficient ω² is positive and constant for the ideal oscillator.

If x is positive, acceleration is negative. If x is negative, acceleration is positive.

The minus sign therefore means toward equilibrium, not “opposite to velocity.” While moving toward equilibrium, the object accelerates in its direction of motion.

Why does the object cross the centre?

A restoring force pushes or pulls the displaced object back. The object gains speed while approaching equilibrium.

At equilibrium, the restoring force vanishes, but the object still has velocity. Inertia carries the object past the centre.

The force then reverses direction and slows the object. At a turning point, the object briefly stops and starts moving inward again.

Without energy losses, this exchange continues. A restoring force alone does not guarantee perpetual motion when friction removes energy.

• Amplitude A: the greatest displacement magnitude from equilibrium, measured in metres for linear motion.
• Period T: the time for one complete cycle, measured in seconds.
• Frequency f: the number of cycles per second, measured in hertz (Hz).
• Angular frequency ω: the rate of phase change, measured in radians per second.

Their relationships are:

f = 1/T

ω = 2πf = 2π/T

T = 2π/ω

Frequency and angular frequency describe the same repetition using different scales. One cycle corresponds to 2π radians, not one radian.

What counts as a complete cycle?

Start at the right turning point. Travel to the left turning point, then return to the right turning point.

That complete journey takes T. Travelling between the two turning points takes T/2.

The distance travelled during one complete linear SHM cycle is 4A. The displacement after that complete cycle is zero.

Suppose an oscillator completes 20 cycles in 10 s. Then f = 20/10 = 2 Hz, and T = 0.5 s.

For ideal one-dimensional SHM, compare the centre with the turning points:

The object’s speed is:

speed = ω√(A² − x²)

Its acceleration magnitude is |a| = ω²|x|. As displacement magnitude grows, speed falls while acceleration magnitude increases.

Does “maximum acceleration” mean the signed value?

Not always. At x = +A, a = −ω²A, the most negative acceleration.

At x = −A, a = +ω²A, the most positive acceleration. Both turning points have the same maximum acceleration magnitude.

Near equilibrium, an object can move fast while its acceleration is small. Speed measures motion; acceleration measures how velocity changes.

For a pendulum, the table describes the acceleration component along the swing. The bob’s full acceleration also includes a component toward the pivot.

A convenient expression for ideal SHM is:

x(t) = A cos(ωt + φ)

Here, t is time and φ is the initial phase. Phase identifies the oscillator’s place in its repeating cycle.

Two oscillators can have equal amplitudes and frequencies but different starting positions. Their initial phases distinguish their starting states.

For a release from x = +A at rest, choose φ = 0. The positions then follow this sequence:

Passing the same position does not necessarily complete a cycle. At the two centre crossings, the velocity directions are opposite.

The complete state repeats after T, 2T, 3T, and subsequent whole multiples of T. T is the shortest positive repeat interval for a nonzero SHM oscillation.

Imagine viewing uniform circular motion edge-on. Its projection along a diameter follows SHM, although the rotating object itself follows a circle.

Consider a mass m on a horizontal spring with negligible friction. Within its Hookean range, the spring’s restoring force is:

F = −kx

The spring constant k measures stiffness in newtons per metre (N/m). Here, x measures extension or compression from the relaxed equilibrium length.

Using F = ma gives a = −(k/m)x. Comparing this with a = −ω²x gives:

ω = √(k/m)

T = 2π√(m/k)

A larger mass responds more slowly to the same restoring force. A stiffer spring produces a stronger restoring force for the same displacement.

Therefore, increasing mass lengthens the period. Increasing stiffness shortens the period.

What happens if the mass becomes four times larger?

Keep the same spring and the same ideal conditions. Then T_new/T_old = √4 = 2.

The period doubles; it does not quadruple. If stiffness instead becomes four times larger, the period halves.

Amplitude does not change this ideal Hookean period. Large extensions can violate the linear spring law, so real systems need appropriate limits.

A simple pendulum models a small bob on a light, inextensible string with a fixed pivot. Its length L runs from the pivot to the bob’s centre.

Gravity supplies the restoring component along the bob’s curved path. The tangential force is F_t = −mg sin θ.

For small angles measured in radians, sin θ ≈ θ. This approximation makes the restoring relationship approximately linear.

The small-angle period is:

T ≈ 2π√(L/g)

Here, g is local gravitational acceleration in m/s². The formula assumes small swings, negligible losses, and the simple-pendulum model.

Mass does not appear in this period. Increasing length makes the pendulum slower; increasing g makes it faster.

If length becomes four times larger, the small-angle period doubles. Large swings introduce amplitude dependence, so the small-angle shortcut cannot be applied without checking conditions.

Is the bob’s full acceleration zero at the bottom?

No. The tangential component is zero there, but the moving bob needs centripetal acceleration v²/L toward the pivot.

The bob has maximum speed at the bottom and zero speed at its turning points. Distinguish that speed statement from claims about the full acceleration vector.

For the ideal horizontal spring oscillator, choose potential energy zero at equilibrium. Then:

U = ½kx²

K = ½mv²

E = K + U = ½kA²

U is elastic potential energy, K is kinetic energy, and E is total mechanical energy. Their SI unit is the joule (J).

At equilibrium, potential energy is zero by our chosen reference. Kinetic energy is then maximum.

At either turning point, speed and kinetic energy are zero. Potential energy is then maximum.

Between those positions, energy shifts between kinetic and potential forms. Total mechanical energy stays constant only when the model has no energy losses or external driving.

Potential energy’s zero is a reference choice, not a universal property of every system. Its changes determine the energy exchange.

What happens halfway to a turning point?

At x = A/2, potential energy is E/4 because displacement is squared. Kinetic energy is therefore 3E/4, not E/2.

If amplitude doubles while k stays fixed, E becomes four times larger. The period remains unchanged for the same ideal spring and mass.

Damping removes mechanical energy through processes such as air resistance and friction. With weak damping, an oscillator keeps swinging while its amplitude decreases.

Very strong damping can prevent back-and-forth motion. Returning toward equilibrium does not always mean that oscillations occur.

Forced oscillation means an external repeating influence drives the system. A parent pushing a swing supplies energy through repeated pushes.

Resonance is a strong response when the driving frequency approaches a suitable natural frequency. For a lightly damped oscillator, this occurs near its undamped natural frequency.

Damping limits the response and affects the resonance peak. A real swing does not gain unlimited amplitude merely because the pushes are well timed.

Tuning-fork vibrations and a guitar string’s individual vibration modes can approximately follow harmonic motion. A plucked string may combine several modes, rather than one perfect sinusoidal motion everywhere.

The National Defence Academy and Naval Academy (NDA & NA) examination includes questions about oscillation formulas and conditions.

Acceleration and speed: NDA & NA (II) 2016, Q105

At equilibrium, ideal linear SHM has maximum speed and zero acceleration magnitude. Do not confuse the magnitude with the most negative signed acceleration.

Read Q105.

Recognising the law: NDA & NA (II) 2016, Q106

An equation a_x = −3x has the required restoring linear form. If SI units are used, the coefficient is 3 s⁻².

Thus ω = √3 rad/s, and T = 2π/√3 s. An equation a = +3x pushes away from equilibrium rather than restoring the object.

An equation such as a = −3x² does not have the required proportionality. Squaring x also loses the sign change needed across equilibrium.

Read Q106.

Equal numerical readings: NDA & NA (II) 2016, Q110

Take the stated amplitude A = 2 cm and displacement magnitude |x| = 1 cm. Speed and acceleration have different physical units.

The question does not specify units for the two rates. This solution interprets the numerical readings in cm/s and cm/s², respectively.

Let w represent the numerical angular-frequency value in rad/s. The speed reading in cm/s is w√3.

The acceleration-magnitude reading in cm/s² is w². Equating those numerical readings gives w√3 = w².

For a nonzero oscillator, w = √3. Therefore T = 2π/√3 seconds.

In physical units, the comparison means speed = |acceleration| × 1 s. Writing speed equal to acceleration without a time factor is dimensionally incomplete.

Read Q110.

Repeated cycles: NDA & NA (I) 2018, Q52

The motion does not repeat just once after T. The complete state repeats after every whole multiple: T, 2T, 3T, and so on.

Passing equilibrium after T/2 can give the opposite velocity direction. Position alone does not identify the complete state.

Read the pendulum question.