AP Physics 1 Unit 7 Study Notes: Oscillations and Simple Harmonic Motion

Oscillations, equilibrium, and the idea of a restoring force

An oscillation is a repeated back-and-forth motion about a special position called equilibrium. In equilibrium, the net force on the object is zero, so if the object is exactly at that position and not moving, it will stay there. Oscillations happen when (1) the object is displaced from equilibrium and (2) the forces that result tend to pull or push it back toward equilibrium.

The key physical idea is the restoring force: a force that points toward equilibrium whenever the object is displaced away from equilibrium. If you stretch a spring to the right, the spring pulls left; if you swing a pendulum to the right, gravity’s component along the arc pulls it back left.

Stable vs unstable equilibrium (why “restoring” matters)

Not every equilibrium produces oscillations. What matters is stability:

• Stable equilibrium: a small displacement produces a restoring force back toward equilibrium. This is what you need for oscillations.
• Unstable equilibrium: a small displacement produces a force away from equilibrium (it “runs away” instead of returning).

A classic mental picture is a ball in a bowl (stable) versus a ball balanced on top of a hill (unstable). Oscillations require that “bowl-like” behavior near equilibrium.

Key timing and motion vocabulary

Oscillations are described by a few timing quantities:

• Period: the time for one complete cycle, meaning the motion repeats itself. If you start at maximum right displacement, one period later you are again at maximum right displacement moving in the same direction as before.
• Frequency: the number of cycles per second.

The relationship between them is

f = \frac{1}{T}

where f is frequency (Hz) and T is period (s).

Physics often uses angular frequency, which is a way to measure how fast the “phase” of the motion advances:

\omega = 2\pi f

and therefore

\omega = \frac{2\pi}{T}

Here \omega is in rad/s.

Two more ideas help you describe “how big” and “where in the cycle” the oscillator is:

• Amplitude: the maximum displacement from equilibrium, usually written A.
• Phase: a way to label where you are in the cycle. It is common to use a phase constant \phi when writing a sinusoidal model.

What makes motion “simple harmonic”?

Simple harmonic motion (SHM) is a special kind of oscillation where the restoring force is proportional to displacement and points toward equilibrium. In one dimension, that means the force has the form

F = -kx

where x is displacement from equilibrium and k is a proportionality constant. The negative sign captures “restoring”: if x is positive (to the right), F is negative (to the left).

Why SHM matters: it is one of the most common and useful models in physics. Many real systems behave like SHM for small displacements, even if they are not perfectly sinusoidal for large displacements. That “small displacement” idea shows up strongly for pendulums.

How position, velocity, and acceleration relate in SHM

A defining feature of SHM is that the acceleration points toward equilibrium and grows with displacement. Combining F = -kx with Newton’s second law F = ma gives

a = -\frac{k}{m}x

This relationship is the “signature” of SHM: acceleration is proportional to x and opposite in direction.

Even without calculus, you can reason out important timing relationships:

• At maximum displacement (at x = \pm A), the object momentarily turns around, so velocity is zero.
• At equilibrium (at x = 0), the restoring force is zero, so acceleration is zero, and speed is maximum.

These facts are frequently tested using graphs and qualitative reasoning.

Exam Focus

• Typical question patterns:
  • Identify whether a described system can be modeled as SHM by checking whether the restoring force is proportional to displacement and opposite in direction.
  • Use conceptual checkpoints (at x = 0, acceleration is zero; at x = \pm A, velocity is zero) to interpret or sketch graphs.
  • Convert between T, f, and \omega and interpret what changing each means physically.
• Common mistakes:
  • Mixing up frequency and angular frequency (using 2\pi incorrectly). Always connect them through \omega = 2\pi f.
  • Forgetting that in SHM the acceleration is largest in magnitude at maximum displacement, not at equilibrium.
  • Treating amplitude as “distance traveled in a cycle.” Amplitude is a maximum displacement from equilibrium, not a path length.

Mass-spring systems and Hooke’s law

A mass-spring system is the standard SHM model: a mass attached to an ideal spring on a frictionless surface, or a mass hanging from a vertical spring. The spring provides a restoring force described by Hooke’s law.

Hooke’s law (what it is and what it means)

For an ideal spring, the spring force is proportional to the amount the spring is stretched or compressed relative to its natural (unstretched) length:

F_s = -kx

• F_s is the spring force on the mass.
• k is the spring constant (N/m), measuring stiffness.
• x is the displacement from equilibrium (in SHM modeling) or from natural length (when you first describe stretching).

A higher k means a stiffer spring: for the same displacement, the force magnitude is larger.

A common confusion is what x is measured from. For SHM, it is most convenient to define x = 0 at the equilibrium position of the mass (where net force is zero). Then the force is restoring about that equilibrium.

Why the motion becomes sinusoidal

When Hooke’s law applies and friction is negligible, Newton’s second law gives

ma = -kx

or

a = -\frac{k}{m}x

This specific acceleration rule produces sinusoidal motion in time. In AP Physics 1 you are typically allowed to use the known SHM solution form

x(t) = A\cos(\omega t + \phi)

with

\omega = \sqrt{\frac{k}{m}}

This equation tells you something powerful: if you know the system’s k and m, you know how quickly it oscillates.

Period of a mass-spring oscillator

From \omega = 2\pi/T and \omega = \sqrt{k/m}, you get the mass-spring period:

T = 2\pi\sqrt{\frac{m}{k}}

Interpretation matters as much as the formula:

• Increasing m makes the oscillator “slower” (bigger T).
• Increasing k makes it “faster” (smaller T).
• The period does not depend on amplitude for an ideal spring in SHM.

That last point is a common conceptual target: if you pull the mass back farther (larger A), it travels farther, but it also moves faster in a way that keeps the period the same.

Horizontal vs vertical springs (equilibrium shift)

For a horizontal mass-spring system, equilibrium is typically at the spring’s natural length (if no other horizontal forces act).

For a vertical mass on a spring, gravity matters. At rest, the spring stretches until the upward spring force balances weight:

k\Delta L = mg

where \Delta L is the static stretch from the spring’s natural length.

If you then oscillate the mass up and down, the motion is still SHM about the equilibrium position, not about the natural length. Gravity shifts where equilibrium is, but it does not change the oscillation period (assuming the spring is ideal and k is constant).

Worked example: finding the period and angular frequency

A 0.50\ \text{kg} mass is attached to a spring with k = 200\ \text{N/m} on a frictionless surface.

1) Angular frequency:

\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.50}} = \sqrt{400} = 20\ \text{rad/s}

2) Period:

T = \frac{2\pi}{\omega} = \frac{2\pi}{20} = \frac{\pi}{10}\ \text{s} \approx 0.314\ \text{s}

Notice how the numbers connect: a stiffer spring (large k) makes \omega larger, which makes T smaller.

Worked example: vertical equilibrium stretch

A 0.20\ \text{kg} mass hangs at rest from a spring with k = 50\ \text{N/m}. The equilibrium stretch is

\Delta L = \frac{mg}{k} = \frac{0.20\cdot 9.8}{50} = 0.0392\ \text{m}

If you later oscillate it, you measure displacement from that stretched equilibrium position when using SHM formulas.

Exam Focus

• Typical question patterns:
  • Use Hooke’s law and Newton’s second law to justify that the acceleration is proportional to displacement and therefore the motion is SHM.
  • Compute T, f, or \omega for a mass-spring system when given m and k.
  • For vertical springs, find equilibrium stretch using k\Delta L = mg and then describe oscillations about that point.
• Common mistakes:
  • Using the natural length as x = 0 for vertical spring SHM without shifting to equilibrium, which leads to sign errors.
  • Thinking amplitude affects period for a spring (in ideal SHM it does not).
  • Confusing the spring force magnitude kx with the net force (net force is zero at equilibrium even though weight and spring force can both be nonzero in vertical setups).

Energy in simple harmonic motion

SHM is not just about forces and graphs; it is also one of the cleanest examples of mechanical energy shifting between forms while the total stays constant (when nonconservative forces are negligible).

The energy “story” of an oscillator

In a mass-spring system, energy continually trades between:

• Spring potential energy when the spring is stretched or compressed.
• Kinetic energy when the mass moves fastest.

At the turning points x = \pm A, the mass is momentarily at rest, so all the energy is stored in the spring. At equilibrium x = 0, the spring is unstretched relative to equilibrium, so spring potential energy is minimum and speed is maximum.

Spring potential energy

The potential energy stored in an ideal spring stretched or compressed by amount x is

U_s = \frac{1}{2}kx^2

This formula is worth interpreting:

• It depends on x^2, so it is the same for +x and -x.
• Doubling displacement quadruples spring potential energy.

Total energy and its connection to amplitude

For an ideal mass-spring oscillator with amplitude A, the total mechanical energy is constant and equals the spring potential energy at maximum displacement:

E = \frac{1}{2}kA^2

This equation is extremely useful: if you know A and k, you know the total energy without needing time information.

Speed at a given position (energy method)

At any position x, energy conservation says

E = K + U_s

with

K = \frac{1}{2}mv^2

and

U_s = \frac{1}{2}kx^2

Using E = \frac{1}{2}kA^2, you can solve for speed as a function of position:

\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2}kx^2

Multiply by 2 and rearrange:

mv^2 = k\left(A^2 - x^2\right)

So

v = \sqrt{\frac{k}{m}\left(A^2 - x^2\right)}

Using \omega^2 = k/m, this is also written as

v = \omega\sqrt{A^2 - x^2}

This relationship captures the qualitative behavior:

• Speed is maximum when x = 0.
• Speed is zero when x = \pm A.

Energy graphs (what they should look like)

You may be asked to sketch or interpret energy vs time during SHM:

• U_s is largest at turning points and smallest at equilibrium.
• K is largest at equilibrium and zero at turning points.
• Total E is constant (horizontal line) if there is no damping.

A frequent conceptual trap is thinking potential energy is smallest at equilibrium because “the spring is relaxed.” That’s only true if you define equilibrium appropriately. For a vertical spring-mass, gravitational potential energy changes too; however, if you measure displacement from equilibrium and focus on the oscillation energy, the “exchange” picture still works cleanly.

Worked example: speed at a given displacement

A 0.50\ \text{kg} mass on a horizontal spring with k = 200\ \text{N/m} oscillates with amplitude A = 0.10\ \text{m}. Find the speed when the mass is at x = 0.060\ \text{m}.

First compute k/m:

\frac{k}{m} = \frac{200}{0.50} = 400

Now use

v = \sqrt{\frac{k}{m}\left(A^2 - x^2\right)}

Compute inside:

A^2 - x^2 = 0.10^2 - 0.060^2 = 0.0100 - 0.0036 = 0.0064

So

v = \sqrt{400\cdot 0.0064} = \sqrt{2.56} = 1.6\ \text{m/s}

Check for reasonableness: since x is not far from equilibrium, the speed should be fairly high but not maximum.

Worked example: energy and amplitude

A spring with k = 80\ \text{N/m} oscillates with amplitude A = 0.25\ \text{m}. The total energy is

E = \frac{1}{2}kA^2 = \frac{1}{2}(80)(0.25^2) = 40\cdot 0.0625 = 2.5\ \text{J}

If the system is ideal (no damping), that 2.5\ \text{J} stays constant.

Exam Focus

• Typical question patterns:
  • Use energy conservation to relate speed and position in SHM, especially to find v at a particular x.
  • Determine total energy from amplitude using E = \frac{1}{2}kA^2.
  • Interpret energy bar charts or energy-vs-position graphs for oscillators.
• Common mistakes:
  • Plugging in x = A into the speed formula and expecting a nonzero speed. At turning points, speed must be zero.
  • Forgetting that energy depends on A^2, so doubling amplitude quadruples energy.
  • Treating total mechanical energy as constant in situations where damping or friction is explicitly present.

The simple pendulum (small-angle oscillations)

A simple pendulum is a small mass (called a bob) hanging from a light string of length L. When displaced and released, it swings back and forth. The pendulum is a different physical system than a spring, but for small angles its motion is also simple harmonic.

The restoring force for a pendulum

The key idea is that gravity always points downward, but the bob moves along a circular arc. The force that changes the bob’s speed along the arc is the component of gravity tangent to the path.

If the bob makes an angle \theta with the vertical, the tangential component of weight has magnitude

F_t = mg\sin\theta

and it points back toward the equilibrium position (straight down). The sign is restoring: if \theta is positive to one side, the tangential component accelerates it back the other way.

Why “small angles” matter

The exact pendulum motion is not perfectly SHM for large angles because \sin\theta is not proportional to \theta.

For small angles (in radians), you can use the small-angle approximation:

\sin\theta \approx \theta

Then the restoring force becomes approximately proportional to displacement (in angular terms), which is what SHM requires.

This is why AP Physics 1 emphasizes “small oscillations” for pendulums: it’s the condition that makes the SHM model valid.

Period of a simple pendulum

For small angles, the period of a simple pendulum is

T = 2\pi\sqrt{\frac{L}{g}}

Important interpretations:

• The period depends on length L and gravitational field strength g.
• The period does not depend on the bob’s mass.
• For small angles, the period is approximately independent of amplitude.

The independence from mass surprises many students at first. The reason is that while a larger mass increases the gravitational force, it also increases the inertia by the same factor, and these effects cancel in the equations.

What changes when you move the pendulum to another planet?

Because T depends on g, a pendulum clock would run differently where gravity is different.

If g decreases, T increases, meaning the pendulum swings more slowly.

Worked example: changing length

A simple pendulum has length L = 1.0\ \text{m} on Earth where g = 9.8\ \text{m/s}^2.

Its period is

T = 2\pi\sqrt{\frac{1.0}{9.8}}

Compute the square root:

\sqrt{\frac{1.0}{9.8}} \approx \sqrt{0.102} \approx 0.319

So

T \approx 2\pi(0.319) \approx 2.01\ \text{s}

If you quadruple the length to 4.0\ \text{m}, the period becomes

T = 2\pi\sqrt{\frac{4.0}{9.8}} = 2\pi\sqrt{4}\sqrt{\frac{1.0}{9.8}} = 2(2\pi\sqrt{\frac{1.0}{9.8}})

So the period doubles. This square-root relationship (period scales like \sqrt{L}) is a common test point.

Pendulum energy (qualitative)

A pendulum continuously trades energy between:

• Gravitational potential energy (highest at the endpoints)
• Kinetic energy (highest at the bottom)

This is the same “exchange” concept as the spring, but with gravitational potential energy instead of spring potential energy.

Exam Focus

• Typical question patterns:
  • Use T = 2\pi\sqrt{L/g} to predict how changing L or g changes the period.
  • Explain why the small-angle approximation leads to SHM and why large angles break the model.
  • Compare two pendulums (different lengths or different planets) and reason about which has larger period.
• Common mistakes:
  • Claiming the pendulum period depends on mass. In the simple pendulum model, it does not.
  • Using the pendulum period formula for large angles without noting that it is an approximation.
  • Confusing the angle \theta (angular displacement) with arc length; be clear what “displacement” your equation refers to.

Describing SHM with graphs and equations (without calculus)

In AP Physics 1, you are often given a graph or a few initial-condition statements and asked to connect them to an SHM model. You do not need calculus to do this well, but you do need a strong sense of the timing relationships.

The standard sinusoidal model

A common model for SHM position is

x(t) = A\cos(\omega t + \phi)

• A: amplitude
• \omega: angular frequency
• \phi: phase constant, which shifts the graph left/right so the model matches your starting conditions

You might also see

x(t) = A\sin(\omega t + \phi)

Both are valid. What matters is that you choose a form and set \phi to match the situation.

How to determine amplitude and period from a position-time graph

From a graph of x vs t:

• Amplitude A is the maximum vertical distance from equilibrium to a peak.
• Period T is the horizontal distance between two identical points in the cycle (peak to peak, trough to trough, or same-direction zero crossing to same-direction zero crossing).

Then compute

\omega = \frac{2\pi}{T}

Phase relationships between x, v, and a (conceptual)

Even if you are not differentiating, you can use turning-point logic and symmetry.

• When x is at a maximum or minimum, the object is turning around, so v = 0.
• When x = 0, the object is passing through equilibrium, so speed is maximum and a = 0.
• Acceleration points toward equilibrium, so its sign is opposite the sign of x.

This means:

• v is “shifted” in time relative to x by a quarter of a period.
• a is exactly opposite in sign to x at the same time.

In equation form, SHM acceleration can be written as

a = -\omega^2 x

This is extremely useful on the exam because it lets you relate instantaneous acceleration to instantaneous position without finding time explicitly.

Using a = -\omega^2 x to link graphs

Suppose you are given an a vs t graph and told the motion is SHM. You can infer:

• When a = 0, the object is at equilibrium (so x = 0).
• When a is maximum positive, x is maximum negative.
• The period of a(t) is the same as the period of x(t).

Worked example: building an SHM equation from a story

A mass-spring oscillator has amplitude A = 0.080\ \text{m} and period T = 0.50\ \text{s}. At t = 0 the mass is at maximum positive displacement.

1) Find \omega:

\omega = \frac{2\pi}{T} = \frac{2\pi}{0.50} = 4\pi\ \text{rad/s}

2) Choose a function form based on the starting condition.

“At maximum positive displacement at t = 0” matches cosine with zero phase, because \cos(0) = 1.

So

x(t) = 0.080\cos(4\pi t)

That single modeling choice (cosine instead of sine) avoids a common phase mistake.

Worked example: using acceleration-position link

An oscillator has angular frequency \omega = 12\ \text{rad/s}. At some moment, its position is x = -0.050\ \text{m}. The acceleration is

a = -\omega^2 x = -(12^2)(-0.050) = 144\cdot 0.050 = 7.2\ \text{m/s}^2

The positive sign means the acceleration points in the positive direction, which makes sense because the mass is left of equilibrium and should accelerate right.

Exam Focus

• Typical question patterns:
  • Extract A and T from a graph, then compute f and \omega.
  • Use starting conditions (such as “released from rest at maximum displacement”) to pick sine vs cosine and determine \phi.
  • Use a = -\omega^2 x to relate instantaneous acceleration and position from graphs or verbal descriptions.
• Common mistakes:
  • Measuring period between a peak and the next zero crossing (that is T/4, not T).
  • Assuming x = 0 implies v = 0. In SHM, x = 0 is where speed is maximum.
  • Getting signs backward in a = -\omega^2 x. Always check: acceleration should point toward equilibrium.

Comparing oscillators: what determines the period?

It is easy to memorize two period formulas and still miss what the exam often cares about: why the period depends on certain quantities and not others.

Mass-spring vs pendulum: what sets the “slowness”?

For a mass-spring oscillator, the restoring force comes from the spring: stiffer spring means stronger restoring force for the same displacement, so it turns around faster and has a shorter period.

For a pendulum (small angles), the restoring effect comes from gravity’s component along the arc. A longer pendulum has a gentler curvature and smaller restoring acceleration for the same angle, so it swings more slowly.

Period dependence table

A useful memory aid is to focus on what provides the restoring influence:

• Spring system: “springiness” k
• Pendulum: gravity g and geometry L

Worked example: ranking periods

You have three oscillators:

• A: mass-spring with m = 1.0\ \text{kg} and k = 100\ \text{N/m}
• B: mass-spring with m = 2.0\ \text{kg} and k = 100\ \text{N/m}
• C: pendulum with L = 1.0\ \text{m} (Earth)

Compute A and B periods relative to each other:

T \propto \sqrt{m} when k is fixed.

So

\frac{T_B}{T_A} = \sqrt{\frac{2.0}{1.0}} = \sqrt{2}

B has the larger period.

For C, you can estimate T_C from earlier as about 2.0\ \text{s}.

For A:

T_A = 2\pi\sqrt{\frac{1.0}{100}} = 2\pi\sqrt{0.01} = 2\pi(0.1) \approx 0.63\ \text{s}

So the ranking is T_C > T_B > T_A.

The key skill here is not arithmetic; it is knowing which variables matter in each model.

Exam Focus

• Typical question patterns:
  • Rank or compare periods when one parameter changes (double m, quadruple L, etc.) without necessarily computing full numbers.
  • Explain verbally why amplitude does not affect period for ideal SHM (spring and small-angle pendulum).
  • Choose the correct period model given a physical description.
• Common mistakes:
  • Applying the spring period formula to a pendulum or vice versa.
  • Saying “bigger amplitude means longer period” as a general rule. That is not true for ideal SHM; it can become true for non-ideal pendulums at large angles.
  • Forgetting that the pendulum formula assumes small angles and a simple pendulum setup.

Damping, driving, and resonance (what changes in real life?)

Up to now, the discussion has assumed an “ideal” oscillator: no friction, no air resistance, no internal energy losses. Real oscillators lose energy, and many are also pushed by external periodic forces. AP Physics 1 typically treats these ideas qualitatively: you should understand the physical effects and interpret graphs, even if you do not do advanced differential-equation math.

Damping: losing energy over time

Damping means the oscillator experiences a nonconservative force (like friction or air drag) that removes mechanical energy. The most visible effect is that the amplitude decreases with time.

What stays true:

• The oscillator still tends to move back and forth about equilibrium.

What changes:

• The total mechanical energy decreases over time.
• The amplitude decreases.
• The period may change slightly depending on the damping strength, but in many classroom contexts the main focus is the decaying amplitude and energy loss.

A good way to reason: if energy decreases, then the maximum speed at equilibrium must decrease too, because maximum speed is tied to total energy.

Driven oscillations: adding energy with an external force

A driven oscillator has an external periodic force applied, like pushing a swing repeatedly. The key idea is energy transfer: the driver can add energy to offset damping (or even increase amplitude if damping is small).

Resonance: when driving matches the natural tendency

Resonance occurs when the driving frequency is near the system’s natural frequency (the frequency it would have if left alone). When the timing matches, each push adds energy efficiently, and the amplitude can become large.

A crucial conceptual point: resonance is not “any time you push.” It is about pushing with the right timing.

Real-world relevance:

• Pushing a playground swing: you push once per cycle at the right moments.
• Structural vibrations: buildings and bridges can be damaged if external periodic forces (wind, earthquakes, machinery) excite resonance.

Phase in driven systems (conceptual)

At low driving frequencies, the system’s motion can stay roughly “in step” with the driving force. Near resonance and above it, the timing (phase relationship) between the driver and the motion can shift. You are not usually asked to calculate that phase shift in AP Physics 1, but you may be asked to describe that resonance involves the driver adding energy at the right times.

Exam Focus

• Typical question patterns:
  • Interpret a graph where amplitude decreases with time and connect it to damping and energy loss.
  • Explain qualitatively why resonance produces large amplitudes and why that can be dangerous or useful.
  • Describe how a driven oscillator can reach a steady amplitude when energy added per cycle balances energy lost to damping.
• Common mistakes:
  • Claiming that in damping the frequency must always decrease a lot; the most direct, testable effect is decreasing amplitude and energy.
  • Thinking resonance means “maximum frequency.” Resonance is about matching the system’s natural frequency.
  • Ignoring energy: resonance is fundamentally efficient energy transfer from driver to oscillator.

Experimental and data-analysis skills with oscillations

Oscillations are especially friendly to experiments because periods are measurable with a stopwatch and because the underlying relationships can be linearized to extract parameters like k or g.

Measuring a period well

A single cycle can be hard to time precisely because human reaction time is a significant fraction of a second. A common lab strategy is:

• Time N cycles (like 10 or 20 cycles).
• Compute T by dividing.

If t_N is the time for N cycles, then

T = \frac{t_N}{N}

This reduces the percent error from reaction time.

Finding a spring constant k from period measurements

For a mass-spring oscillator,

T = 2\pi\sqrt{\frac{m}{k}}

Square both sides:

T^2 = \frac{4\pi^2}{k}m

This has the form of a line: T^2 is proportional to m.

If you graph T^2 (vertical axis) vs m (horizontal axis), the slope is

\text{slope} = \frac{4\pi^2}{k}

So

k = \frac{4\pi^2}{\text{slope}}

Why this matters: AP questions often ask you to interpret a graph and identify the physical meaning of slope.

Finding g from pendulum measurements

For a simple pendulum (small angle),

T = 2\pi\sqrt{\frac{L}{g}}

Square:

T^2 = \frac{4\pi^2}{g}L

Graph T^2 vs L:

\text{slope} = \frac{4\pi^2}{g}

So

g = \frac{4\pi^2}{\text{slope}}

Worked example: extracting k from a line

A student plots T^2 vs m for a spring and gets a best-fit slope of 0.80\ \text{s}^2/\text{kg}.

Use

k = \frac{4\pi^2}{\text{slope}}

Compute:

k = \frac{4\pi^2}{0.80}

Since 4\pi^2 \approx 39.5,

k \approx \frac{39.5}{0.80} \approx 49\ \text{N/m}

The result is in N/m because the algebraic units work out from the linearized equation.

What experimental “gotchas” look like

Oscillation labs often go wrong in predictable ways:

• Pendulum angle too large: violates the small-angle approximation, causing period to differ from the simple formula.
• Timing only one cycle: increases uncertainty.
• Spring not ideal: large stretches can make Hooke’s law less accurate.

On exam-style questions, these show up as prompts like “A student’s measured period is larger than predicted. Give a plausible reason.” You should connect the reason to a specific violated assumption.

Exam Focus

• Typical question patterns:
  • Use linearized equations (like T^2 vs m or T^2 vs L) to interpret slope and solve for k or g.
  • Design or critique an experiment to reduce uncertainty (time many cycles, keep angles small, repeat trials).
  • Reason about how non-ideal conditions change results relative to ideal predictions.
• Common mistakes:
  • Plotting T vs m and expecting a straight line; the linear relationship is between T^2 and m.
  • Forgetting units on slope and therefore getting unrealistic parameter values.
  • Treating disagreement with theory as “random error” when it is actually systematic (for example, using large-angle pendulum swings).