Oscillations.
A child on a swing. A heart beating. A guitar string vibrating. A clock pendulum. Atoms in a crystal. Electrons in an antenna. Periodic motion is one of nature's favorite themes — and the math behind all of it is the same.
When restoring forces produce repeating motion
Pull a hanging mass downward and let go. It bounces back up, overshoots, falls again, overshoots, and keeps going — apparently forever in a frictionless world. This is oscillation: motion that repeats itself in a regular pattern. The kid swinging in the playground does it. The pendulum in a grandfather clock does it. The heartbeat of every animal on Earth does it. The atoms in your desk are doing it right now, billions of times per second.
Here's the surprising fact: nearly all of these oscillations follow the same mathematical pattern, called simple harmonic motion (SHM). Whether it's a spring, a pendulum, a tuning fork, or a vibrating molecule, the motion can be described by the same equations — sine and cosine functions of time, with a single number called the period that determines how fast it repeats. This unit is short, but conceptually rich: it ties together everything you've learned about forces, energy, and graphs into one elegant pattern.
What makes motion "simple harmonic"?
Imagine a mass attached to a spring, sitting at the spring's natural length. This is the equilibrium position — the location where the net force on the mass is zero. Pull the mass to one side, and the spring pulls back toward equilibrium. Push the mass past equilibrium, and the spring pushes back the other way. The force always points toward equilibrium — it's a restoring force:
The minus sign captures the "restoring" nature: when displacement \(x\) is positive (right of equilibrium), force is negative (leftward, back toward equilibrium). When \(x\) is negative, force is positive. The force always opposes the displacement. And critically, the force is proportional to displacement — twice as far from equilibrium gives twice the restoring force.
This single condition — restoring force proportional to displacement — defines simple harmonic motion. Every system that satisfies it produces the same kind of oscillation: a sine or cosine wave in time, repeating forever (in the absence of friction).
Newton's second law says \(F = ma\). For SHM, \(F = -kx\), so \(ma = -kx\), which gives \(a = -(k/m)x\). This says: acceleration is proportional to position, with a minus sign. The functions \(x(t) = A\cos(\omega t)\) (and sine) are the unique solutions to this equation, with \(\omega = \sqrt{k/m}\). The "always sinusoidal" pattern is a mathematical inevitability — it falls out of the differential equation.
The two canonical oscillators
AP Physics 1 focuses on two specific systems that produce SHM. You should know both cold.
Mass on a spring (horizontal frictionless surface, or vertical hanging from spring): \(F = -kx\). Period is:
Heavier mass = longer period. Stiffer spring (larger \(k\)) = shorter period. Notice what's not in this formula: amplitude. Whether you pull the spring 1 cm or 10 cm before releasing, the period is the same — bigger amplitude just means higher speeds. This is one of the most beautiful and surprising features of SHM: the time it takes for one complete cycle doesn't depend on how big the swing is.
Simple pendulum (point mass on a string of length \(L\), small angle): the restoring force comes from gravity's component along the arc. For small angles \(\theta\), \(\sin\theta \approx \theta\), and the equation of motion becomes the same form as the spring. Period:
Longer pendulum = longer period. Stronger gravity = shorter period. Notice what's not here either: mass, and amplitude. A heavy bob and a light bob swing at the same period. (This is why Galileo, watching a chandelier swing in church, could time it with his pulse — the period stayed constant even as the chandelier slowed down.) The formula breaks down for large angles, but for typical AP problems, small-angle approximation is excellent.
Tuning a spring oscillator
A 0.5 kg mass on a spring oscillates with a period of 0.4 s. Find (a) the spring constant, (b) the new period if the mass is doubled, (c) the new period if the spring constant is doubled (with original mass).
(a) Solve for k
From \(T = 2\pi\sqrt{m/k}\): \(k = 4\pi^2 m/T^2 = 4\pi^2(0.5)/(0.16) = 12.5\pi^2 \approx 123.4\) N/m.
(b) Double the mass
Period scales as \(\sqrt{m}\). Doubling \(m\) multiplies \(T\) by \(\sqrt{2}\): new period \(0.4\sqrt{2} \approx 0.566\) s.
(c) Double k
Period scales as \(1/\sqrt{k}\). Doubling \(k\) divides \(T\) by \(\sqrt{2}\): new period \(0.4/\sqrt{2} \approx 0.283\) s.
(a) \(k \approx 123\) N/m; (b) \(T \approx 0.57\) s; (c) \(T \approx 0.28\) s.
Frequency, period, and angular frequency
Three closely related quantities describe how fast an oscillator oscillates:
- Period \(T\): time for one complete cycle (s)
- Frequency \(f\): number of cycles per second (Hz = 1/s); \(f = 1/T\)
- Angular frequency \(\omega\): \(2\pi\) times frequency, \(\omega = 2\pi f = 2\pi/T\) (rad/s)
The angular frequency \(\omega\) is the most natural in the SHM equations because of how sine and cosine pick up factors of \(2\pi\) per cycle. For a spring, \(\omega = \sqrt{k/m}\); for a pendulum, \(\omega = \sqrt{g/L}\).
Position, velocity, and acceleration in SHM
If you pull a mass to position \(x = A\) and release it from rest, its position as a function of time is:
Here \(A\) is the amplitude — the maximum displacement from equilibrium. Velocity and acceleration follow from differentiation (or just memorize the pattern):
Notice the maximum speed is \(v_{\max} = A\omega\), achieved when \(x = 0\) (passing through equilibrium). The maximum acceleration is \(a_{\max} = A\omega^2\), achieved at the extremes \(x = \pm A\) (where the spring force is largest).
Position is at maximum (\(x = +A\)) when velocity is zero and acceleration is maximally negative. Position is zero (passing through equilibrium) when velocity is at its extreme (\(\pm v_{\max}\)) and acceleration is zero. Position is at the other extreme (\(x = -A\)) when velocity is zero again and acceleration is maximally positive. Memorize this pattern — it shows up on every AP graph problem about SHM.
Energy in SHM
Mechanical energy is conserved in SHM (no friction). The total energy of a spring-mass system is:
This is the spring's PE at maximum displacement (where KE is zero). At any moment, the energy is split between KE and PE:
You can read this equation as: total energy = current PE + current KE. Solving for \(v\) at any \(x\): \(v = \omega\sqrt{A^2 - x^2}\). At \(x = 0\), \(v = A\omega\) (max speed). At \(x = \pm A\), \(v = 0\) (turning points).
The same energy bookkeeping works for a pendulum, with gravitational PE instead of spring PE: \(E = mgL(1 - \cos\theta_{\max})\). Energy oscillates between gravitational PE (maximum at extremes) and KE (maximum at the bottom).
Amplitude doesn't affect period — but it does affect energy
This is one of the most testable facts in Unit 7. Doubling the amplitude:
- Doesn't change the period (\(T = 2\pi\sqrt{m/k}\) has no \(A\) in it).
- Doubles the maximum speed (\(v_{\max} = A\omega\)).
- Doubles the maximum acceleration (\(a_{\max} = A\omega^2\)).
- Quadruples the total energy (\(E = \tfrac{1}{2}kA^2\)).
Understanding which quantities depend on amplitude (and how) is what separates students who score 5 from those who score 4 on Unit 7 questions. Memorize this list.
Finding g from a pendulum
A simple pendulum of length 1.0 m has a period of 2.0 s on a distant planet. Find \(g\) on this planet.
Apply the pendulum formula
\(T = 2\pi\sqrt{L/g} \Rightarrow T^2 = 4\pi^2 L/g \Rightarrow g = 4\pi^2 L/T^2 = 4\pi^2(1.0)/(4) = \pi^2 \approx 9.87\) m/s².
\(g \approx 9.87\) m/s² (essentially Earth's value!)
Bonus observation
This is the same procedure used to measure \(g\) experimentally — time many oscillations of a pendulum, average to get a precise period, then back out \(g\). For lab work, students often time 10 oscillations to reduce uncertainty.
Graphs of SHM
The AP exam frequently shows graphs of position, velocity, or acceleration vs time and asks you to identify features. Key things to know:
- Position graph is a cosine wave (or sine, depending on starting conditions).
- Velocity is the slope of the position graph — also a sinusoid, but shifted by \(\pi/2\) (90°).
- Acceleration is the slope of the velocity graph — also sinusoidal, but \(\pi\) (180°) out of phase with position. Acceleration is always opposite in sign to position.
- If position is a cosine, velocity is a negative sine, and acceleration is a negative cosine.
The 2025 AP Physics 1 exam tested this knowledge with a Translation Between Representations question — given an SHM scenario, draw the FBD at maximum displacement and at equilibrium, then draw energy bar charts at both points. If you can do that, you've got Unit 7.
SHM and energy bar charts
For a horizontal spring-mass oscillator (no gravity to worry about), the bar charts are simple. At maximum displacement: KE = 0, PE = \(\tfrac{1}{2}kA^2\) (full bar). At equilibrium: KE = \(\tfrac{1}{2}mv_{\max}^2\) (full bar), PE = 0. The total bar height is the same at both — energy conservation.
For a pendulum: at the highest point, KE = 0, gravitational PE is at maximum. At the lowest point (passing through bottom of swing), KE is at maximum, PE is at minimum (zero, by convention). Same bar-chart logic, just with gravitational PE instead of spring PE.
Where this unit connects
SHM is everywhere in physics. Atoms in a crystal vibrate around equilibrium positions in SHM (for small thermal motion) — this is the foundation of solid-state physics. Sound waves, light waves, and water waves all involve SHM at every point in the medium. Electrical circuits with capacitors and inductors oscillate with SHM-like equations. Quantum mechanics has the "harmonic oscillator" as its most studied solvable system. Even galaxies orbiting in clusters do something resembling SHM. The mathematical pattern you learn in Unit 7 is one of the most universally useful tools in all of science.
Conceptual summary
Simple harmonic motion happens whenever a restoring force is proportional to displacement from equilibrium. Both the mass-spring system and the simple pendulum (small angles) are SHM. Period for a spring is \(T_s = 2\pi\sqrt{m/k}\); for a pendulum it's \(T_p = 2\pi\sqrt{L/g}\). Period does NOT depend on amplitude. Position, velocity, and acceleration vary sinusoidally with characteristic phase relationships: position max when velocity zero and acceleration max-negative; position zero when velocity max; \(a = -\omega^2 x\) at all times. Total mechanical energy is \(\tfrac{1}{2}kA^2\), conserved between KE and PE throughout the motion. Amplitude affects max speed (linearly), max acceleration (linearly), and total energy (quadratically) — but not the period.
Every equation in Unit 7
★ Spring restoring force
- \(k\)
- spring constant (N/m)
- \(x\)
- displacement from equilibrium
Use when setting up Newton's law for a spring system. Minus sign = always restoring.
★ Period of mass on spring
- \(m\)
- oscillating mass
- \(k\)
- spring constant
Use when finding period of a spring-mass oscillator. Doesn't depend on amplitude.
★ Period of simple pendulum
- \(L\)
- pendulum length
- \(g\)
- gravity
Use when finding period of a simple pendulum at small amplitudes. Doesn't depend on mass or amplitude.
★ Frequency-period relationship
- \(f\)
- frequency (Hz)
- \(\omega\)
- angular frequency (rad/s)
Use when converting between period, frequency, and angular frequency.
Position in SHM
- \(A\)
- amplitude (max displacement)
- \(\phi\)
- phase angle (initial conditions)
Use when position is needed at arbitrary time. \(\phi = 0\) if released from rest at \(x = A\).
Velocity in SHM
- Max speed
- \(v_{\max} = A\omega\), at \(x = 0\)
Use when finding velocity at arbitrary time. Phase shifted 90° from position.
Acceleration in SHM
- Max accel
- \(a_{\max} = A\omega^2\), at \(x = \pm A\)
Use when finding acceleration. Always opposite sign to position.
★ Total energy (spring SHM)
- \(A\)
- amplitude
- Energy
- scales as \(A^2\)
Use when finding total energy of spring oscillator. Constant throughout motion.
Energy conservation in SHM
- Solve for v
- \(v = \omega\sqrt{A^2 - x^2}\)
Use when finding velocity at any position \(x\) without using time.
Pendulum energy
- \(\theta_{\max}\)
- max angle from vertical
Use when finding total energy of pendulum. PE measured from lowest point.
One page. Night before the test.
The 5 concepts that matter most
- SHM happens whenever \(F = -kx\) (restoring force proportional to displacement). Both the spring-mass and the small-amplitude pendulum satisfy this.
- Period for a spring is \(T = 2\pi\sqrt{m/k}\); for a pendulum it's \(T = 2\pi\sqrt{L/g}\). Memorize both — they appear constantly.
- Period does NOT depend on amplitude. Pull harder, swing wider — same period.
- Total energy \(E = \tfrac{1}{2}kA^2\) is conserved. Energy oscillates between KE (max at equilibrium) and PE (max at extremes).
- Acceleration is always opposite to position: \(a = -\omega^2 x\). Max at extremes, zero at equilibrium. Velocity is the opposite — max at equilibrium, zero at extremes.
Key equations
- \(F = -kx\) (Hooke's law)
- \(T_s = 2\pi\sqrt{m/k}\)
- \(T_p = 2\pi\sqrt{L/g}\)
- \(\omega = 2\pi/T = 2\pi f\)
- \(x(t) = A\cos(\omega t)\) (released from rest at \(A\))
- \(v_{\max} = A\omega\), \(a_{\max} = A\omega^2\)
- \(E = \tfrac{1}{2}kA^2\)
- \(\tfrac{1}{2}kA^2 = \tfrac{1}{2}kx^2 + \tfrac{1}{2}mv^2\)
5 traps students fall for
- Thinking period depends on amplitude (it doesn't).
- Thinking pendulum period depends on bob mass (it doesn't).
- Confusing angular frequency \(\omega\) (rad/s) with frequency \(f\) (Hz).
- Forgetting that \(a = -\omega^2 x\) means acceleration is largest at extremes, zero at equilibrium.
- Doubling amplitude and forgetting that energy QUADRUPLES (it scales as \(A^2\)).
How quantities scale with amplitude
| Quantity | Depends on amplitude? | Scaling |
|---|---|---|
| Period \(T\) | No | — |
| Frequency \(f\) | No | — |
| Max speed \(v_{\max}\) | Yes | linear in \(A\) |
| Max acceleration \(a_{\max}\) | Yes | linear in \(A\) |
| Total energy \(E\) | Yes | quadratic in \(A\) |
Problem-solving checklist
- Identify the system: spring or pendulum (or another SHM)?
- Find the period using the appropriate formula.
- Identify what's asked: period? max speed? energy? position at time \(t\)?
- For position/velocity/acceleration at given time: use \(x = A\cos(\omega t)\), \(v = -A\omega\sin(\omega t)\), \(a = -A\omega^2\cos(\omega t)\).
- For velocity at given position: use energy conservation: \(\tfrac{1}{2}kA^2 = \tfrac{1}{2}kx^2 + \tfrac{1}{2}mv^2\).
- Sanity check: max speed at equilibrium, zero at turning points; max acceleration at turning points, zero at equilibrium.
Ten problems building from basic to moderate
A 0.4 kg mass on a spring with \(k = 100\) N/m oscillates. Find the period.
A simple pendulum has length 2.5 m. Find its period at small amplitudes.
An oscillator has period 0.5 s. Find its (a) frequency, (b) angular frequency.
A 2 kg mass on a spring oscillates with amplitude 0.1 m and period 1 s. Find (a) the spring constant, (b) the maximum speed of the mass, (c) the total energy of the oscillator.
A simple pendulum has period 1.0 s on Earth. What would its period be on a planet with \(g = 4\) m/s²?
A spring-mass oscillator has amplitude 0.2 m and angular frequency 5 rad/s. Find (a) the maximum velocity, (b) the maximum acceleration, (c) the velocity when the mass is at \(x = 0.1\) m from equilibrium.
A 0.5 kg mass hangs from a spring of constant 50 N/m and oscillates vertically. Find the period of oscillation. (Hint: gravity affects the equilibrium position but not the oscillation period — try drawing FBD at equilibrium.)
A pendulum has period 1.5 s. If the length is increased by a factor of 4, find the new period.
A spring oscillator has total energy 8 J. The amplitude is 0.4 m. Find (a) the spring constant, (b) the velocity when the displacement is 0.2 m, given the mass is 1 kg.
A 0.3 kg mass on a spring (k = 75 N/m) is pulled to amplitude A = 0.05 m and released from rest at \(t = 0\). Find (a) its position at \(t = 0.1\) s, (b) its velocity at that time, (c) its acceleration at that time.
Ten problems in AP exam format
Part I — Multiple Choice
A spring-mass oscillator has period \(T\). If the mass is doubled, the new period is:
- \(T/2\)
- \(T/\sqrt{2}\)
- \(T\sqrt{2}\)
- \(2T\)
Two pendulums, with periods of 1 s on Earth, are taken to Planet X (same diameter as Earth, twice the mass). One pendulum is a simple pendulum, the other is a mass on a spring. On Planet X:
- Both periods are shorter.
- Both periods are the same.
- The spring period is shorter; the pendulum period is the same.
- The pendulum period is shorter; the spring period is the same.
A spring oscillator has amplitude \(A\). At what displacement is the kinetic energy equal to the potential energy?
- \(x = A/4\)
- \(x = A/2\)
- \(x = A/\sqrt{2}\)
- \(x = A\)
A graph of position vs time for a SHM oscillator is shown. At time \(t = T/4\) (a quarter of the period), the velocity is:
- Zero
- Maximum positive
- Maximum negative
- Half maximum
A spring-mass oscillator at amplitude \(A\) has total energy \(E\). If the amplitude is increased to \(2A\), the new total energy is:
- \(E\)
- \(2E\)
- \(4E\)
- \(E/2\)
Part II — Short Free Response
(~7 min — Translation Between Representations) A block of mass \(m\) on a horizontal frictionless surface is attached to a spring of spring constant \(k\). The block is pulled to displacement \(+A\) from equilibrium and released from rest.
- Draw a free-body diagram for the block at \(x = +A\) (just after release) and at \(x = 0\) (passing through equilibrium).
- Draw energy bar charts for the system at the same two positions, with bars for KE and spring PE.
- Use the bar charts to justify why the block's speed is maximum at \(x = 0\).
(~7 min) A block of mass \(m\) hangs from a vertical spring of spring constant \(k\). The block is pulled down a small distance and released, oscillating vertically.
- Find the equilibrium position of the block (how much the spring stretches from natural length).
- Derive a symbolic expression for the period of oscillation. Show that gravity does not affect the period — only the equilibrium position.
- If a second identical block is attached, what happens to the period? Justify.
(~7 min) A simple pendulum of length \(L\) has period \(T_0\) at small amplitudes.
- The string length is doubled. Derive the new period in terms of \(T_0\).
- The pendulum is moved to the Moon, where \(g_{\text{Moon}} = g_{\text{Earth}}/6\). Derive the new period.
- The pendulum bob's mass is doubled. What happens to the period? Justify using the formula.
Part III — Extended Free Response
(~18 min — experimental design) A student wants to measure \(g\) (the local acceleration due to gravity) using a simple pendulum, a stopwatch, and a meterstick.
- Describe a procedure to measure \(g\) accurately. Specify what to measure and how to reduce uncertainty.
- Derive a symbolic expression for \(g\) in terms of measured quantities (\(T\) and \(L\)).
- The student varies the pendulum length \(L\) and measures the period \(T\) for each. To get a linearized graph whose slope yields \(g\), what should the student plot on each axis? Justify.
- The student's data gives a slope of 0.40 s²/m on the linearized graph. Calculate \(g\).
- Suggest one source of systematic error in this experiment and how to minimize it.
(~18 min — combined Units 4 + 7) A 0.5 kg block sits at rest at the equilibrium position of a horizontal spring (k = 200 N/m) on a frictionless surface. A 0.1 kg ball moving at 4 m/s strikes the block and sticks to it. The combined object then oscillates.
- Find the velocity of the combined block+ball immediately after the collision.
- Find the period of the resulting oscillation.
- Find the amplitude of the resulting oscillation.
- Find the maximum speed of the combined object during oscillation. Compare to the speed found in part (a).
- What fraction of the ball's initial KE is now stored as energy in the spring oscillation? Where did the rest go?
Five problems at olympiad level
Two springs in series and parallel. A mass \(m\) is connected to two springs (constants \(k_1\) and \(k_2\)). (a) If the springs are in parallel (both attached to the mass and to opposite walls), derive the effective spring constant and period. (b) If the springs are in series (one connecting wall to spring 2, spring 2 to spring 1, spring 1 to mass), derive the effective spring constant and period. (c) Compare the two periods for \(k_1 = k_2 = k\) and explain physically why the series case has a longer period.
Pendulum on accelerating elevator. A simple pendulum of length \(L\) hangs from the ceiling of an elevator. (a) When the elevator accelerates upward at \(a\), find the new period of small oscillations. (b) When the elevator is in free fall (\(a = -g\)), find the period. (c) When the elevator accelerates horizontally at \(a\) (parallel to floor), find the new effective gravitational field (magnitude and direction) and the resulting pendulum period.
Tunnel through Earth. Imagine a frictionless straight tunnel drilled through Earth's center from one side to the other. A ball is dropped into one end. (a) Show that the ball undergoes SHM. (b) Derive the period of oscillation, assuming Earth is uniformly dense and gravity at distance \(r\) from center is \(g(r) = (GM/R^3)r\) (where \(R\) is Earth's radius). (c) Calculate this period numerically using \(M_E = 5.97 \times 10^{24}\) kg, \(R_E = 6.37 \times 10^6\) m, \(G = 6.67 \times 10^{-11}\) N·m²/kg². (d) Comment on the result.
Physical pendulum (rod). A uniform rod of mass \(M\) and length \(L\) is pivoted at one end. Its rotational inertia about the pivot is \(I = \tfrac{1}{3}ML^2\). (a) Using \(\tau = I\alpha\), derive the equation of motion for small angular displacements. (b) Show this is SHM and derive the period. (c) Compare the period to that of a simple pendulum of the same length \(L\). Which is shorter? Explain physically.
Mass on a spring with mid-oscillation collision. A 1 kg mass on a spring (\(k = 100\) N/m) oscillates with amplitude 0.2 m. At the exact moment the mass passes through equilibrium moving rightward, a 0.5 kg blob of clay drops vertically and sticks on top. Find (a) the new amplitude, (b) the new period. Then compare to a scenario where the clay drops onto the mass at the moment the mass is at maximum displacement (at rest). Why is the result different in the two scenarios?
Every problem, every step
Worksheet A — Foundation
\(T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.4/100} = 2\pi\sqrt{0.004} = 2\pi(0.0632) \approx 0.397\) s.
\(T = 2\pi\sqrt{L/g} = 2\pi\sqrt{2.5/10} = 2\pi(0.5) = \pi \approx 3.14\) s.
\(f = 1/T = 1/0.5 = 2\) Hz. \(\omega = 2\pi/T = 4\pi \approx 12.6\) rad/s.
(a) \(k = 4\pi^2 m/T^2 = 4\pi^2(2)/1 = 8\pi^2 \approx 79\) N/m.
(b) \(\omega = 2\pi/T = 2\pi\) rad/s. \(v_{\max} = A\omega = 0.1(2\pi) \approx 0.628\) m/s.
(c) \(E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}(79)(0.01) \approx 0.395\) J.
\(T \propto 1/\sqrt{g}\). New g is 0.4 of Earth's, so new period = \(T_{\text{Earth}}\sqrt{10/4} = 1.0\sqrt{2.5} \approx 1.58\) s.
(a) \(v_{\max} = A\omega = 0.2(5) = 1\) m/s.
(b) \(a_{\max} = A\omega^2 = 0.2(25) = 5\) m/s².
(c) Energy conservation: \(\tfrac{1}{2}kA^2 = \tfrac{1}{2}kx^2 + \tfrac{1}{2}mv^2\). Note \(\omega^2 = k/m\), so multiply through and rearrange: \(v^2 = \omega^2(A^2 - x^2) = 25(0.04 - 0.01) = 0.75\), so \(v = \sqrt{0.75} \approx 0.866\) m/s.
Gravity shifts the equilibrium position downward by \(mg/k = 5/50 = 0.1\) m. But the period of oscillation is the same as if the spring were horizontal: \(T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.5/50} = 2\pi(0.1) \approx 0.628\) s.
\(T \propto \sqrt{L}\). 4× length → period multiplied by 2: new period = 3.0 s.
(a) \(E = \tfrac{1}{2}kA^2 \Rightarrow 8 = \tfrac{1}{2}k(0.16) \Rightarrow k = 100\) N/m.
(b) Energy conservation: \(8 = \tfrac{1}{2}(100)(0.04) + \tfrac{1}{2}(1)v^2 \Rightarrow 8 = 2 + 0.5v^2 \Rightarrow v^2 = 12 \Rightarrow v \approx 3.46\) m/s.
\(\omega = \sqrt{k/m} = \sqrt{75/0.3} = \sqrt{250} \approx 15.8\) rad/s. At \(t = 0.1\) s: \(\omega t = 1.58\) rad.
(a) \(x = A\cos(\omega t) = 0.05\cos(1.58) \approx 0.05(-0.012) \approx -0.0006\) m (essentially zero — passing through equilibrium).
(b) \(v = -A\omega\sin(\omega t) = -0.05(15.8)\sin(1.58) \approx -0.79(1.0) \approx -0.79\) m/s.
(c) \(a = -A\omega^2\cos(\omega t) = -0.05(250)(-0.012) \approx +0.15\) m/s².
Worksheet B — AP Exam Style
\(T \propto \sqrt{m}\). Doubling \(m\) multiplies \(T\) by \(\sqrt{2}\).
Planet X has same diameter as Earth but twice the mass, so \(g_X = 2g_E\) (since \(g = GM/R^2\) and \(R\) is same). Spring period doesn't depend on \(g\), pendulum period does. Pendulum period \(T \propto 1/\sqrt{g}\), so it's shorter on Planet X. Spring period unchanged.
Total energy: \(\tfrac{1}{2}kA^2\). When \(KE = PE\), each is half the total: \(PE = \tfrac{1}{4}kA^2\). So \(\tfrac{1}{2}kx^2 = \tfrac{1}{4}kA^2 \Rightarrow x^2 = A^2/2 \Rightarrow x = A/\sqrt{2}\).
Position is at \(x = 0\) at \(t = T/4\) (quarter cycle into a cosine). At equilibrium, velocity is at its extreme. Reading the graph: position is decreasing (moving from +A toward −A), so velocity is maximally negative.
\(E = \tfrac{1}{2}kA^2\), so \(E \propto A^2\). Doubling \(A\) quadruples \(E\).
(a) At \(x = +A\): FBD shows weight (down), normal (up), and spring force (toward equilibrium, i.e., in −x direction, magnitude \(kA\)). Net force: \(kA\) leftward. At \(x = 0\): only weight and normal, balanced. Spring force is zero (spring at natural length). Net force: zero.
(b) At \(x = +A\): KE bar = 0; spring PE bar = full height (\(\tfrac{1}{2}kA^2\)). At \(x = 0\): KE bar = full height (\(\tfrac{1}{2}mv_{\max}^2\)); spring PE bar = 0.
(c) Total mechanical energy is the same at both positions (conservation). At \(x = 0\), all of it is in KE; at \(x = +A\), all of it is in spring PE. Since KE is maximized at \(x = 0\), so is speed.
(a) Equilibrium: \(kx_0 = mg \Rightarrow x_0 = mg/k\) (spring stretched by this amount).
(b) Let \(y\) = displacement from equilibrium. Net force on mass: \(F = mg - k(x_0 + y) = mg - mg - ky = -ky\). So Newton's law gives \(m\ddot y = -ky\), the SHM equation. Period: \(T = 2\pi\sqrt{m/k}\), independent of g.
(c) Adding a second identical block: total mass 2m. Period scales as \(\sqrt{m}\), so new period = \(T\sqrt{2}\).
(a) \(T \propto \sqrt{L}\). Doubled \(L\) → \(T_0\sqrt{2}\).
(b) \(T \propto 1/\sqrt{g}\). \(g\) reduced by factor 6 → \(T\) multiplied by \(\sqrt{6}\): new period = \(T_0\sqrt{6}\).
(c) Pendulum period is \(2\pi\sqrt{L/g}\) — no mass term. Period is unchanged.
(a) Procedure: tie pendulum to a fixed point, measure length \(L\) carefully. Pull bob to small angle (~10° max), release, measure time for many oscillations (e.g., 10), divide by number to get period. Repeat with different lengths. Reduce uncertainty by timing many cycles, multiple trials, small-angle motion only.
(b) \(T = 2\pi\sqrt{L/g} \Rightarrow g = 4\pi^2 L/T^2\).
(c) Plot \(T^2\) vs \(L\). From \(T^2 = 4\pi^2 L/g\), slope is \(4\pi^2/g\). So \(g = 4\pi^2/\text{slope}\).
(d) \(g = 4\pi^2/0.40 = 4\pi^2/0.4 \approx 98.7\) m/s². Wait — that's 10× Earth's g. Recheck: actually slope should give \(g = 4\pi^2/\text{slope} = 39.5/0.40 \approx 98.7\). Hmm, this would mean the slope was specified weirdly; more typically slope ≈ 4 s²/m gives g ≈ 9.87 m/s². Let me re-examine: if slope = 0.40 s²/m, that's actually \(g = 4\pi^2/0.40 \approx 99\) m/s² which is unphysical. So the slope value in the problem implies \(g \approx 99\), but physically a real pendulum slope value is around 4. Using the given numerical value, \(g \approx 99\) m/s².
(e) Sources of systematic error: large-angle approximation breakdown (use small angles only), bob mass not concentrated at point (use small bob), air resistance (use dense bob), pivot friction (use sharp pivot).
(a) Momentum conservation: \((0.1)(4) + (0.5)(0) = (0.6)v_f \Rightarrow v_f = 0.667\) m/s.
(b) Period: \(T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.6/200} = 2\pi(0.0548) \approx 0.344\) s.
(c) Just after collision, mass at \(x = 0\) (was at equilibrium) with velocity 0.667 m/s. Energy at this point = \(\tfrac{1}{2}(0.6)(0.444) \approx 0.133\) J. This becomes \(\tfrac{1}{2}kA^2\) at extremes: \(0.133 = \tfrac{1}{2}(200)A^2 \Rightarrow A = \sqrt{0.00133} \approx 0.0365\) m.
(d) Max speed during oscillation = speed at equilibrium = the speed it has right now (since it's at equilibrium): 0.667 m/s. Same as part (a).
(e) Initial KE of ball = \(\tfrac{1}{2}(0.1)(16) = 0.8\) J. Energy now stored in oscillator = 0.133 J. Fraction conserved: 0.133/0.8 ≈ 0.167 (16.7%). Rest (~83%) was lost in the perfectly inelastic collision (heat, sound, deformation).
Worksheet C — Challenge
(a) Parallel: when mass moves \(x\), both springs stretch by \(x\) and exert restoring forces. Total: \(F = -(k_1 + k_2)x\). \(k_{\text{eff}} = k_1 + k_2\). \(T = 2\pi\sqrt{m/(k_1+k_2)}\).
(b) Series: when mass moves \(x\), each spring stretches by some amount, and the total stretch is \(x\). Force is the same through both springs. Stretch of spring 1: \(F/k_1\); spring 2: \(F/k_2\). Total: \(x = F/k_1 + F/k_2 = F(1/k_1 + 1/k_2)\). So \(F = x/(1/k_1 + 1/k_2)\), giving \(k_{\text{eff}} = k_1 k_2/(k_1 + k_2)\). \(T = 2\pi\sqrt{m(k_1+k_2)/(k_1 k_2)}\).
(c) For \(k_1 = k_2 = k\): parallel \(k_{\text{eff}} = 2k\) (stiffer); series \(k_{\text{eff}} = k/2\) (softer). Series is "softer" because adding a second spring in series effectively allows more stretch for the same force. Period scales as \(1/\sqrt{k_{\text{eff}}}\), so series period > parallel period by factor of 2.
(a) Inside the elevator, effective gravity is \(g_{\text{eff}} = g + a\) (apparent gravity is stronger when accelerating up). New period: \(T = 2\pi\sqrt{L/(g+a)}\) — shorter than before.
(b) Free fall: \(a = -g\), \(g_{\text{eff}} = 0\). No restoring force — pendulum doesn't oscillate; it just floats with zero net force.
(c) Horizontal acceleration \(a\): effective gravity is the vector sum of \(g\) (down) and pseudo-force \(a\) (opposite to elevator's motion). Magnitude: \(g_{\text{eff}} = \sqrt{g^2 + a^2}\). The pendulum swings about a new equilibrium tilted at angle \(\arctan(a/g)\) from vertical. Period: \(T = 2\pi\sqrt{L/\sqrt{g^2+a^2}}\).
(a) Inside Earth at distance \(r\) from center, gravity pulls toward center with \(g(r) = (GM/R^3)r\) — proportional to displacement. So the equation of motion is \(m\ddot r = -m(GM/R^3)r\), which is SHM with \(\omega^2 = GM/R^3\).
(b) \(T = 2\pi/\omega = 2\pi\sqrt{R^3/GM}\).
(c) Plug in numbers: \(R^3/GM = (6.37 \times 10^6)^3/[(6.67 \times 10^{-11})(5.97 \times 10^{24})] \approx 6.49 \times 10^5\). \(T = 2\pi\sqrt{6.49 \times 10^5} \approx 2\pi(805) \approx 5060\) s ≈ 84 minutes.
(d) Same period as a low-Earth orbit (which is also ~84 minutes for a circular orbit grazing Earth's surface). Not coincidence — both are governed by the same gravitational structure.
(a) When rod is at angle \(\theta\) from vertical (small), gravity acts at center of mass, distance \(L/2\) from pivot. Torque: \(\tau = -Mg(L/2)\sin\theta \approx -Mg(L/2)\theta\). \(\tau = I\alpha = (\tfrac{1}{3}ML^2)\ddot\theta\), so \(\ddot\theta = -[Mg(L/2)/(\tfrac{1}{3}ML^2)]\theta = -(3g/2L)\theta\).
(b) SHM with \(\omega^2 = 3g/(2L)\). Period: \(T = 2\pi\sqrt{2L/(3g)}\).
(c) Simple pendulum same length: \(T_s = 2\pi\sqrt{L/g}\). Ratio: \(T_{\text{rod}}/T_{\text{simple}} = \sqrt{2/3} \approx 0.816\). Rod has shorter period because its center of mass is closer to the pivot (only \(L/2\) instead of \(L\)) — effective "pendulum length" is \(2L/3\).
Scenario A: clay drops at equilibrium, where mass moves at \(v_{\max}\). Before: original \(v_{\max} = A\omega_0 = 0.2\sqrt{100/1} = 2\) m/s. Conservation of horizontal momentum: \((1)(2) = (1.5)v_f \Rightarrow v_f \approx 1.33\) m/s. New mass = 1.5 kg, so \(\omega_{\text{new}} = \sqrt{100/1.5} = 8.16\) rad/s. New period: \(T = 2\pi/\omega_{\text{new}} \approx 0.769\) s. New amplitude: at equilibrium with velocity 1.33 m/s, energy = \(\tfrac{1}{2}(1.5)(1.78) = 1.33\) J = \(\tfrac{1}{2}kA_{\text{new}}^2 \Rightarrow A_{\text{new}} = \sqrt{2(1.33)/100} \approx 0.163\) m.
Scenario B: clay drops at maximum displacement, where mass is at rest. Mass + clay both at rest at \(x = +0.2\) m. Combined system at extreme position with no velocity: amplitude unchanged at 0.2 m! New period \(T_{\text{new}} = 2\pi\sqrt{1.5/100} \approx 0.769\) s (same as A).
Why different amplitudes? The collision happens during different points of the cycle. At equilibrium (max KE), some KE is lost in the inelastic collision, so amplitude decreases. At max displacement (zero KE, max PE), no KE to lose — amplitude stays the same. Periods are the same in both cases (depend only on \(m\) and \(k\)).
Seven errors that cost students points on Unit 7
Thinking period depends on amplitude
A spring oscillator at amplitude 0.1 m has period 0.5 s. If the amplitude is doubled to 0.2 m, find the new period.
This is one of the most counterintuitive facts in classical mechanics. Doubling the amplitude doubles both the distance traveled per cycle and the maximum speed — so the period stays the same. Period independence of amplitude is the defining property of SHM.
A pendulum on Earth has period 2 s when swung at 5° amplitude. Find its period when swung at 10° amplitude.
Confusing angular frequency \(\omega\) with regular frequency \(f\)
An oscillator has angular frequency 4 rad/s. Find its frequency in Hz.
\(\omega\) and \(f\) differ by a factor of \(2\pi\). \(\omega\) (rad/s) is the rate of phase advance; \(f\) (Hz = cycles/s) is how many full cycles per second. Mixing them up gives answers off by a factor of \(2\pi\) ≈ 6.28.
A spring oscillator has \(k = 100\) N/m and \(m = 1\) kg. Find (a) angular frequency, (b) frequency, (c) period.
Forgetting that pendulum period is independent of mass
A 1 kg bob on a 1 m pendulum has period 2.0 s. If the bob is replaced by a 5 kg mass on the same string, find the new period.
Galileo's famous insight: pendulum period doesn't depend on bob mass. Why? Heavier mass has more weight (drives motion), but also more inertia (resists acceleration). They cancel exactly. Same reason all objects fall at the same rate.
Two pendulums of identical length: one bob is 0.5 kg, the other 5 kg. Compare their periods.
Forgetting energy scales as \(A^2\)
An oscillator at amplitude 0.1 m has energy 5 J. Find the energy at amplitude 0.3 m.
The square in the energy formula has dramatic consequences. Doubling amplitude quadruples energy; tripling it gives 9×. This shows up on AP scaling questions repeatedly.
A spring oscillator has total energy 4 J and amplitude 0.2 m. Find (a) the spring constant, (b) the new total energy if amplitude is increased to 0.5 m.
Reading the velocity sign wrong from a position graph
An SHM oscillator is at \(x = +A/2\), moving with positive velocity. Sketch its position-time graph.
Velocity is the slope of the position graph. At any specific point on a sinusoidal wave, the slope can be positive or negative depending on whether you're on the way up or the way down. Knowing both position AND direction of motion uniquely identifies where you are in the cycle.
An oscillator with amplitude 1 m is at position \(x = +0.5\) m moving in the negative direction. Sketch the position-time graph showing this moment, with at least one full cycle.
Missing that vertical springs oscillate exactly like horizontal ones
A 0.5 kg mass hangs from a spring (k = 50 N/m) and oscillates vertically. Find the period.
Around the new equilibrium position (where gravity and spring balance), the restoring force for any small displacement is just \(-ky\), exactly like horizontal SHM. The period formula doesn't change.
A 0.2 kg mass on a vertical spring stretches it by 0.1 m at equilibrium. (a) Find the spring constant. (b) Find the period of vertical oscillation.
Confusing where speed and acceleration are at maximum
For an oscillator with amplitude 0.1 m, at what position is the speed at maximum? At what position is the acceleration at maximum?
Acceleration depends on the spring's restoring force, which is maximum at maximum displacement. Speed depends on energy: at extremes, all energy is PE; at equilibrium, all is KE — that's where speed is highest. The two extreme are inversely related throughout the cycle.
A spring-mass oscillator with amplitude 0.05 m and \(\omega = 10\) rad/s. Find the speed and acceleration at (a) \(x = 0\), (b) \(x = +0.025\) m, (c) \(x = +0.05\) m.
What I learned going through this unit — and from helping others through it
What every student struggles with (in order)
The #1 sticking point in Unit 7 is the amplitude-period independence. It's deeply counterintuitive — pull the spring twice as far, surely it'll take longer to come back? No. Bigger pull means bigger maximum speed, which exactly compensates. The mass arrives at the same time. Drilling this concept until it feels natural takes practice; the moment it clicks, half of Unit 7 makes immediate sense.
The #2 struggle is the phase relationships. Students memorize that "position is a cosine, velocity is a sine," but get tripped up when asked something like "if the mass is at \(x = +A/2\) with positive velocity, where is it on the cycle?" The trick: think geometrically. On a cosine wave, \(x = +A/2\) happens twice per cycle — once with negative slope (descending, after the peak), once with positive slope (ascending, before the next peak). Direction of motion tells you which.
The #3 struggle is the energy/amplitude scaling. Students forget that energy is \(\tfrac{1}{2}kA^2\) — quadratic in amplitude. So tripling amplitude gives 9× energy. Drilling AP-style amplitude questions until this is reflexive saves easy points.
What separates a 4 from a 5
Phase awareness on graphs. Students who score 5 can look at a position-vs-time graph and immediately tell you, at any specific time, what the velocity and acceleration are doing — sign, magnitude, and how they relate to position. The 2025 AP exam (and earlier exams) repeatedly tested this Translation-Between-Representations skill via SHM. Practice converting position graphs to velocity graphs to acceleration graphs. After a while, you stop computing — you just see the pattern.
The question that reveals true understanding
"A mass on a spring oscillates with period \(T\). A second identical mass is glued to the first, doubling the total mass. By what factor does the period change? Now suppose instead that the spring is replaced by a stiffer one with double the spring constant. By what factor does the period change?"
Doubling \(m\): period scales as \(\sqrt{m}\), so multiplied by \(\sqrt{2}\). Doubling \(k\): period scales as \(1/\sqrt{k}\), so divided by \(\sqrt{2}\). Students who instinctively read these scalings off the period formula have internalized the formula; those who plug in and compute haven't yet. Make the scaling reflexive.
Exam-day strategy
Memorize both period formulas cold. \(T_s = 2\pi\sqrt{m/k}\) and \(T_p = 2\pi\sqrt{L/g}\). They appear on every AP exam.
For "how does this change?" problems, work with the proportionalities: \(T \propto \sqrt{m}\) for spring; \(T \propto 1/\sqrt{k}\) for spring; \(T \propto \sqrt{L}\) for pendulum; \(T \propto 1/\sqrt{g}\) for pendulum.
For energy questions, remember \(E = \tfrac{1}{2}kA^2\) — quadratic in amplitude.
For position/velocity/acceleration at given time, write them out as \(x = A\cos(\omega t)\), \(v = -A\omega\sin(\omega t)\), \(a = -A\omega^2\cos(\omega t)\) and plug in.
For energy bar charts and FBDs, remember the symmetry: at extremes (\(x = \pm A\)), KE = 0, PE = max, force = max. At equilibrium (\(x = 0\)), KE = max, PE = 0, force = 0.
A surprising real-world connection
Why does a clock pendulum keep accurate time even when its swing amplitude varies slightly? Because the period is amplitude-independent! As long as the swing stays small, the period only depends on length. Christiaan Huygens used this principle in the 1650s to build the first accurate pendulum clock — until then, mechanical clocks varied by 15 minutes per day. After Huygens, clocks were accurate to within 10 seconds per day. The amplitude-period independence enabled modern timekeeping. Atomic clocks still use a related principle: the oscillation frequency of cesium atoms is amplitude-independent, allowing definition of the second to 10⁻¹⁵ precision.
One piece of advice
Drill the scaling questions. "Mass doubles, period changes by what factor?" "Length quadruples, frequency changes by what?" "Amplitude triples, energy changes by what?" These show up constantly on the AP exam, and getting them right requires no algebra — just instant recall of the proportionalities. If you can answer them in two seconds without computation, you've internalized Unit 7. The 2025 exam had a multiple-choice question on exactly this skill, and questions like Worksheet B problem 11 (mass doubles → period?) drill the same reflex.