Simple Harmonic Motion from a Restoring Force
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You should be comfortable with Newton’s second law, basic calculus (differentiation and integration), and the concept of a potential energy function. Familiarity with trigonometric identities for sine and cosine is also assumed.
Why this matters
Simple Harmonic Motion (SHM) is the fundamental building block of all wave phenomena.
Almost every physical system that oscillates, from a pendulum to an electron in an atom, behaves like a mass on a spring when displaced slightly from equilibrium. By mastering the mathematics of SHM, you gain the tools to understand how energy is stored and transferred, how waves propagate through media, and why frequencies are the natural language of physics. This lecture establishes the core equation of motion and the universal frequency-amplitude independence that defines this idealized motion.
The Intuitive Picture
Imagine a ball resting at the bottom of a smooth, symmetric bowl. If you nudge it slightly to the left, gravity pulls it back toward the center. As it moves, it gains speed, overshoots the center, and then slows down as it climbs the other side. It never stops; it simply reverses direction at the turning points. This back-and-forth motion is periodic. The key insight is that for small displacements, the force pulling the ball back is directly proportional to how far it is from the center. Double the displacement, and you double the restoring force. This linear relationship is what makes the motion "simple" and predictable, resulting in a perfect sinusoidal path in time rather than a chaotic or irregular one.
The Linear Restoring Force
Consider a particle of mass \(m\) moving in one dimension under the influence of a potential energy function \(U(x)\). A stable equilibrium exists at \(x_0\) where the potential has a local minimum. For small displacements \(x\) from this minimum, we can approximate the potential using a Taylor expansion. Since the first derivative vanishes at the minimum (\(U'(x_0) = 0\)), the leading non-constant term is quadratic.
The force is given by \(F = -\frac{dU}{dx}\). For a quadratic potential, this force is linear in displacement. We define the spring constant \(k\) as the second derivative of the potential at the equilibrium point, \(k \equiv U''(x_0)\). Since the equilibrium is stable, \(k\) must be positive. The equation of motion is therefore:
This is the defining equation of Simple Harmonic Motion. It states that the acceleration is proportional to the negative of the displacement. The negative sign indicates that the force always acts to restore the particle to the equilibrium position.
The Sinusoidal Solution
To solve \(m\ddot{x} + kx = 0\), we look for a function whose second derivative is proportional to itself with a negative sign. The sine and cosine functions possess exactly this property. The general solution is a linear combination of these two functions, which can be compactly written as a single cosine with a phase shift:
Here, \(A\) is the amplitude, representing the maximum displacement from equilibrium. \(\varphi\) is the phase constant, determined by the initial position and velocity. The term \(\omega\) is the angular frequency. Substituting this solution back into the equation of motion allows us to determine \(\omega\). The second derivative of \(x(t)\) is \(-\omega^2 A\cos(\omega t + \varphi)\). Equating this to \(-\frac{k}{m}x(t)\) yields the fundamental result:
The period \(T\), the time for one complete cycle, is related to the angular frequency by \(T = \frac{2\pi}{\omega}\). Thus:
Amplitude Independence
A striking feature of the harmonic oscillator is that the frequency \(\omega\) depends only on the mass \(m\) and the stiffness \(k\). It is completely independent of the amplitude \(A\). This means that whether you pull the mass back by 1 cm or 10 cm, it will oscillate at exactly the same rate. This is a unique property of linear systems. In non-linear systems, the frequency often changes with amplitude, leading to complex behaviors. The amplitude independence is a direct consequence of the linearity of the restoring force. It allows us to superpose different oscillations, a principle that becomes crucial when we later discuss Fourier decomposition and wave interference.
The derivation
Simple Harmonic Motion from a Linear Restoring Force
For a particle of mass \(m\) confined near a stable equilibrium of a smooth one-dimensional potential \(U(x)\), small displacements \(x\) from the minimum obey \(m\ddot{x} = -kx\) with \(k \equiv U''(x_0) > 0\), and the motion is sinusoidal, \(x(t) = A\cos(\omega t + \varphi)\), with a single angular frequency \(\omega = \sqrt{k/m}\) that is independent of amplitude.
Do not confuse angular frequency \(\omega\) with ordinary frequency \(f\). Angular frequency is measured in radians per second, while ordinary frequency is measured in Hertz (cycles per second). They are related by \(\omega = 2\pi f\). A common error is to use \(f\) in the equation of motion \(x(t) = A\cos(\omega t + \varphi)\), which would result in incorrect units and a wrong period. Always ensure that the argument of the trigonometric function is dimensionless, which requires \(\omega\) to have units of \(\mathrm{rad/s}\). C
Worked example
Problem. A mass of \(2.0\ \mathrm{kg}\) is attached to a spring with a spring constant of \(50\ \mathrm{N/m}\). The mass is displaced \(0.10\ \mathrm{m}\) from its equilibrium position and released from rest. Determine the angular frequency, the period of oscillation, and the equation of motion.
- Identify the given values: \(m = 2.0\ \mathrm{kg}\), \(k = 50\ \mathrm{N/m}\), \(x(0) = 0.10\ \mathrm{m}\), \(v(0) = 0\ \mathrm{m/s}\).
- Calculate the angular frequency using \(\omega = \sqrt{k/m}\): \[ \omega = \sqrt{\frac{50}{2.0}} = \sqrt{25} = 5.0\ \mathrm{rad/s} \]
- Calculate the period using \(T = \frac{2\pi}{\omega}\): \[ T = \frac{2\pi}{5.0} \approx 1.26\ \mathrm{s} \]
- Determine the amplitude and phase. Since the mass is released from rest at maximum displacement, \(A = 0.10\ \mathrm{m}\). At \(t=0\), \(x(0) = A\cos(\varphi) = 0.10\), so \(\cos(\varphi) = 1\), which implies \(\varphi = 0\).
- Write the equation of motion: \[ x(t) = 0.10\cos(5.0t)\ \mathrm{m} \]
Units check. The spring constant \(k\) has units \(\mathrm{N/m} = \mathrm{kg/s^2}\). The ratio \(k/m\) has units \(\mathrm{s^{-2}}\). The square root \(\sqrt{k/m}\) has units \(\mathrm{s^{-1}}\), which is correct for angular frequency. The argument of the cosine, \(\omega t\), has units \(\mathrm{s^{-1} \cdot s} = 1\) (dimensionless), which is required for trigonometric functions.
Three quick tests
- What is the relationship between the period \(T\) and the angular frequency \(\omega\)?
Answer
\(T = \frac{2\pi}{\omega}\). The period is the time for one full cycle, while \(\omega\) is the rate of change of the phase angle. - If the mass \(m\) is quadrupled, how does the period \(T\) change?
Answer
The period doubles. Since \(T \propto \sqrt{m}\), increasing \(m\) by a factor of 4 increases \(T\) by a factor of \(\sqrt{4} = 2\). - Why is the frequency independent of the amplitude in SHM?
Answer
Because the restoring force is linear in displacement. This linearity ensures that the equation of motion is linear, leading to a solution where the frequency is determined solely by the system parameters \(m\) and \(k\), not by the initial conditions (amplitude).
Problem set for PU-103 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.