Mass-spring systems and simple pendulum — AP Physics C: Mechanics
1. Core Overview of Oscillator Systems ★☆☆☆☆ ⏱ 2 min
Mass-spring systems and simple pendulums are the two core physical examples of systems that undergo simple harmonic motion (SHM), the focus of Unit 6 Oscillations in the AP Physics C: Mechanics CED. This subtopic accounts for approximately 6-8% of the total AP exam score, appearing regularly on both multiple choice (MCQ) and free response (FRQ) sections, often combined with energy, force, or differential equation questions.
Mastery of core relationships here is required for all further work on oscillations in AP Physics C: Mechanics.
2. Angular Frequency and Period for Mass-Spring Oscillators ★★☆☆☆ ⏱ 4 min
To find the period of a mass-spring oscillator, we start with Newton's second law for displacements from equilibrium. For any mass-spring system, the restoring force follows Hooke's law:
-kx = ma \implies a = -\left(\frac{k}{m}\right)x
By definition, SHM has acceleration of the form $a = -\omega^2 x$, where $\omega$ is angular frequency. Matching terms gives $\omega = \sqrt{\frac{k}{m}}$, so period:
T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{k}}
A common point of confusion is the effect of gravity for vertical or incline mass-spring systems. Gravity adds a constant force to the mass, which only shifts the equilibrium position of the system. When we measure displacement from this new equilibrium, gravity cancels out of the restoring force, so the same formula for $\omega$ and $T$ holds regardless of orientation.
Exam tip: If a problem gives you the equilibrium stretch of a vertical spring alongside mass and spring constant, the stretch is almost always a distractor for period calculation. You only need $m$ and $k$ to find $T$.
3. The Simple Pendulum and Small-Angle Approximation ★★★☆☆ ⏱ 4 min
A simple pendulum consists of a point mass $m$ on a massless, inextensible string of length $L$ fixed at one end. The restoring force acts tangential to the arc of the pendulum's motion, and is given by:
F = -mg\sin\theta
For SHM, we need a restoring force proportional to displacement. For small angles (typically less than ~10°), the approximation $\sin\theta \approx \theta$ (for $\theta$ in radians) holds, so the restoring force simplifies to:
F \approx -mg\theta
The arc length displacement from equilibrium is $s = L\theta$, so $\theta = \frac{s}{L}$. Substituting gives:
F = -\left(\frac{mg}{L}\right)s
This matches Hooke's law with effective spring constant $k_{\text{eff}} = \frac{mg}{L}$. Using the SHM angular frequency formula $\omega = \sqrt{\frac{k_{\text{eff}}}{m}}$ gives $\omega = \sqrt{\frac{g}{L}}$, so period:
T = 2\pi\sqrt{\frac{L}{g}}
Note that the mass cancels out, so the period of a simple pendulum is independent of the mass of the bob and (for small angles) independent of amplitude.
Exam tip: Always confirm the problem asks for the period of small oscillations before using $T = 2\pi \sqrt{L/g}$. For large angular displacements, the actual period is always longer than the small-angle prediction.
4. Energy Conservation in Undamped Oscillating Systems ★★★☆☆ ⏱ 4 min
For undamped (frictionless) SHM, total mechanical energy is conserved. This relationship is frequently used to solve for speed at a given displacement, faster than integrating the SHM differential equation. For mass-spring systems, total energy is the sum of kinetic energy and elastic potential energy:
E = KE + U = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \text{constant}
At the maximum displacement (turning point), $x = A$ (amplitude) and $v=0$, so all energy is potential: $E = \frac{1}{2}kA^2$. This gives the energy conservation relation:
\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2
For simple pendulums, total energy is the sum of kinetic energy and gravitational potential energy. Taking equilibrium as the zero potential reference, at maximum angular displacement $\theta_{\text{max}}$, the bob rises by $h = L(1-\cos\theta_{\text{max}})$, so maximum potential energy equals total energy.
Exam tip: When asked for speed at a non-equilibrium, non-turning point position, energy conservation is almost always faster and less error-prone than using the explicit SHM position-velocity formula.
Common Pitfalls
Why: You may incorrectly assume gravity changes the restoring force, when it only shifts the equilibrium position of the oscillator.
Why: You transfer the mass dependence from mass-spring systems to pendulums without re-deriving the relationship.
Why: The small-angle approximation comes from the Taylor series of $\sin\theta$, which is only valid when $\theta$ is in radians.
Why: You confuse Hooke's force $F=kx$ with elastic potential energy $U=\frac{1}{2}kx^2$.
Why: You forget that amplitude independence is only an approximation that holds for small angles where $\sin\theta \approx \theta$.