AP Physics 1 Unit 7 Oscillations: Understanding SHM Through Representations and Energy

Representing and Analyzing SHM

What makes motion “simple harmonic”?

Simple harmonic motion (SHM) is a special kind of back-and-forth (oscillatory) motion where the object experiences a restoring influence that points back toward equilibrium and grows in proportion to how far you are from equilibrium. In practice, the most important identifying feature is:

• The acceleration (and thus net force) is proportional to displacement and opposite in direction.

Mathematically, the defining relationship is:

a=-\omega^2 x

Here, x is the displacement from equilibrium (positive or negative), a is the acceleration, and \omega is the angular frequency (how “fast” the oscillation cycles, in radians per second). This equation matters because it gives you an immediate test for SHM: if an object’s acceleration-vs-position relationship is a straight line through the origin with negative slope, it’s SHM.

A classic physical cause of SHM is a restoring force that behaves like Hooke’s law:

F=-kx

Combining this with Newton’s 2nd law F=ma gives:

ma=-kx

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

Comparing with a=-\omega^2 x, you get:

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

So for a mass–spring oscillator, the spring constant k and the mass m set the “stiffness vs inertia” balance, which sets the oscillation rate.

Why the “proportional and opposite” requirement matters

If the restoring effect is not proportional to displacement (for example, friction-dominated motion, or large-angle pendulum motion where \sin\theta is not well-approximated), then the motion is not truly SHM: the period may change with amplitude, and the graphs won’t be perfect sinusoids.

Key quantities and how they relate

When you describe SHM, you repeatedly use a small set of parameters. The big idea is that many different representations (equations, graphs, energy bar charts) are all describing the same motion.

Equilibrium position: where net force is zero (often x=0).
Amplitude A: the maximum displacement from equilibrium.
Period T: time for one full cycle.
Frequency f: cycles per second.
Angular frequency \omega: radians per second.

These are connected by:

f=\frac{1}{T}

\omega=2\pi f

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

A very common misconception is to think amplitude affects period for an ideal mass–spring oscillator. In ideal SHM for a mass–spring system, T depends on m and k, not on A.

For a mass–spring system:

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

For a simple pendulum at small angles (the “small-angle approximation,” where the motion becomes SHM), the period is:

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

where L is the pendulum length and g is the gravitational field strength near Earth’s surface.

The sinusoidal model (position, velocity, acceleration)

In SHM, the position as a function of time is sinusoidal. A standard model is:

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

• A sets the maximum displacement.
• \omega sets how quickly it oscillates.
• \phi is the **phase constant**, which shifts where in the cycle you start at t=0.

In AP Physics 1, you typically don’t need calculus to use this model effectively, but it helps to know the standard companion relationships (they can be reasoned from the idea that the motion is sinusoidal and that velocity is largest when position is zero):

v(t)=-A\omega\sin(\omega t+\phi)

a(t)=-A\omega^2\cos(\omega t+\phi)

Notice something powerful: comparing a(t) and x(t),

a(t)=-\omega^2 x(t)

That’s exactly the defining SHM condition again.

Phase relationships (how the graphs “line up”)

A very typical skill is interpreting time graphs:

• x(t) and a(t) are always opposite in sign (they are “out of phase” by half a cycle).
• v(t) is shifted relative to x(t) by a quarter of a cycle.

Instead of memorizing phase shifts, you can reason physically:

• At maximum displacement (at x=\pm A), the object must turn around, so v=0 there.
• At equilibrium (at x=0), the restoring force is zero, so acceleration is zero, and speed is maximum.

From the velocity equation, the maximum speed is:

v_{\max}=A\omega

And the maximum acceleration magnitude is:

a_{\max}=A\omega^2

A common mistake is swapping where maxima occur (for example, claiming acceleration is maximum at equilibrium). In SHM, acceleration magnitude is maximum at the endpoints because the restoring force is largest there.

Using graphs as representations of SHM

AP questions often give you graphs and ask you to extract physics.

1) The a vs x graph: the SHM “fingerprint”

For SHM:

a=-\omega^2 x

So an acceleration-position graph is a straight line through the origin with slope:

\text{slope}=\frac{a}{x}=-\omega^2

That means:

\omega=\sqrt{-\text{slope}}

and then

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

This is extremely useful because it lets you determine the period without knowing the amplitude.

Worked example (from an a vs x graph)
A graph of acceleration vs displacement is a line through the origin with slope -16\ \text{s}^{-2}. Find \omega and T.

1) Use the slope relation:

-\omega^2=-16

\omega^2=16

\omega=4\ \text{rad/s}

2) Convert to period:

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

What can go wrong: Students sometimes use \omega=-4 because the slope is negative. By convention, \omega is taken as a positive magnitude; the negative sign is already accounted for in a=-\omega^2 x.

2) The x vs t graph: reading amplitude and period

From a displacement-time graph:

• A is the maximum vertical distance from equilibrium.
• T is the horizontal distance for one full cycle.

Then:

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

If the graph shows the object starts at a maximum displacement, a cosine model with \phi=0 fits naturally. If it starts at equilibrium moving positive, a sine model or a cosine with a phase shift fits better. In AP Physics 1, you often don’t need to calculate \phi precisely; you mainly need the qualitative starting conditions.

3) The v vs t and a vs t graphs

Once you know x(t) is sinusoidal, the others must be sinusoidal with the same period.

You can use key points:

• Where x(t) is at a max or min, v(t)=0.
• Where x(t)=0, |v(t)| is maximum.
• Where x(t)=0, a(t)=0.
• Where x(t)=\pm A, |a(t)| is maximum.

Connecting different physical systems to the same SHM math

One of the most important “analysis” ideas is that many systems behave like SHM if they have a restoring influence proportional to displacement.

Mass–spring oscillator

Restoring force:

F=-kx

Angular frequency:

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

Period:

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

Small-angle pendulum (why “small-angle” matters)

For a pendulum, the tangential restoring component is:

F_t=-mg\sin\theta

For small angles (in radians), \sin\theta\approx\theta, so:

F_t\approx -mg\theta

Using arc-length displacement x=L\theta, so \theta=\frac{x}{L}:

F_t\approx -mg\frac{x}{L}

This has the Hooke’s-law form F=-k_{\text{eff}}x with:

k_{\text{eff}}=\frac{mg}{L}

So the motion becomes SHM with:

\omega=\sqrt{\frac{g}{L}}

and

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

What can go wrong: If the amplitude is large, \sin\theta\approx\theta is no longer accurate, and the motion’s period increases slightly with amplitude. On exams, if they say “small oscillations” or “small angle,” you’re being told to use the SHM model.

Notation reference (common equivalents)

Exam Focus

• Typical question patterns:
  • Given an a vs x graph (or data), determine whether the motion is SHM and find \omega or T from the slope.
  • From a graph of x(t), read A and T, then find \omega, v_{\max}, or a_{\max}.
  • Compare two oscillators (different m, k, or L) and predict how T changes using proportional reasoning.
• Common mistakes:
  • Treating amplitude as affecting period for an ideal mass–spring oscillator; use T=2\pi\sqrt{\frac{m}{k}} to check dependencies.
  • Mixing up where speed/acceleration are maximum; remember: |v| max at equilibrium, |a| max at endpoints.
  • Using the negative slope of a vs x as a negative \omega; take \omega as a positive magnitude.

Energy of Simple Harmonic Oscillators

Energy as a “second lens” on SHM

SHM can be analyzed using forces and acceleration, but energy gives you a complementary, often simpler perspective: the oscillator trades energy back and forth between kinetic energy and potential energy while total mechanical energy stays constant (if nonconservative forces like friction are negligible).

Energy analysis matters because many exam problems ask for speeds and positions without explicitly involving time. Instead of writing x(t) and differentiating, you can often use conservation of mechanical energy.

The total mechanical energy is:

E=K+U

where K is kinetic energy and U is potential energy.

Spring–mass oscillator energy

For an ideal spring, the elastic potential energy stored when the spring is stretched or compressed by displacement x is:

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

The kinetic energy of the mass when it has speed v is:

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

If there’s no damping, total energy is constant:

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

Where is energy located during the cycle?

• At the endpoints x=\pm A: the mass is momentarily at rest, so v=0.

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

• At equilibrium x=0: the spring is neither stretched nor compressed, so U_s=0 and speed is maximum.

E=\frac{1}{2}mv_{\max}^2

Setting these equal gives:

\frac{1}{2}mv_{\max}^2=\frac{1}{2}kA^2

v_{\max}=A\sqrt{\frac{k}{m}}

Using \omega=\sqrt{\frac{k}{m}}, this becomes the earlier result:

v_{\max}=A\omega

This is a nice example of consistency: the graph/kinematics view and the energy view give the same relationship.

Speed at a particular displacement

From conservation of energy:

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

Solve for v:

\frac{1}{2}mv^2=\frac{1}{2}k(A^2-x^2)

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

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

(You choose the sign based on direction of motion if needed; AP often only asks for speed, the positive magnitude.)

Worked example (speed using energy)
A mass m=0.50\ \text{kg} on a 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}.

1) Use the speed-displacement relation:

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

2) Substitute values:

v=\sqrt{\frac{200}{0.50}(0.10^2-0.060^2)}

3) Compute inside step by step:

\frac{200}{0.50}=400

0.10^2=0.0100

0.060^2=0.0036

A^2-x^2=0.0064

So:

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

What can go wrong: Students sometimes plug in x=A and are surprised they get v=0. That’s correct: at the turning points, all energy is stored in the spring.

Total energy and amplitude

For a spring oscillator, the total mechanical energy is:

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

This is a powerful result: amplitude is essentially an energy measure. Double the amplitude and you quadruple the total energy.

You can also rewrite total energy in terms of m, \omega, and A using k=m\omega^2 (from \omega^2=\frac{k}{m}):

E=\frac{1}{2}m\omega^2 A^2

Common misconception: “Heavier mass means more total energy.” Not necessarily. If you keep the same physical spring and the same amplitude, E=\frac{1}{2}kA^2 does not include m. The mass changes \omega and T, but for fixed k and A, the total energy is set by the spring’s stiffness and how far you stretch/compress it.

Energy graphs in SHM

Energy representations show patterns that are easy to test against physical reasoning.

For a spring oscillator:

• U_s(x)=\frac{1}{2}kx^2 is a parabola opening upward.
• Total energy E=\frac{1}{2}kA^2 is a horizontal line (constant).
• Kinetic energy is:

K(x)=E-U_s(x)=\frac{1}{2}k(A^2-x^2)

So K(x) is also a downward-opening parabola with maximum at x=0.

These shapes help you answer qualitative questions like “Where is the object moving fastest?” (where K is largest) and “Where is the acceleration largest?” (where |x| is largest).

Pendulum energy (using gravitational potential)

A pendulum’s restoring influence comes from gravity. Even if you don’t use the small-angle SHM formula, energy ideas are still straightforward.

Take the lowest point of the swing as zero gravitational potential. At some angle, the bob is higher by a vertical height h, so:

U_g=mgh

As it swings down, gravitational potential converts to kinetic energy:

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

v=\sqrt{2gh}

This is especially useful because it doesn’t require trig-heavy force analysis. In many AP problems, you’ll be given geometry (or the vertical height change directly) so you can compute h.

Connection to SHM: For small oscillations, the pendulum behaves like SHM with a constant period (approximately independent of amplitude). Energy still trades between U_g and K just like the spring trades between elastic potential and kinetic.

Using energy to infer SHM relationships (without time)

Energy can also help you justify common SHM claims:

• At equilibrium, potential energy is minimum, so kinetic energy is maximum, so speed is maximum.
• At turning points, kinetic energy is zero, so speed is zero.
• Larger amplitude means more total energy.

You can also connect energy to acceleration indirectly: since restoring force is larger at larger |x|, the acceleration magnitude is larger at larger |x|, which matches the energy idea that the potential energy curve is steepest at larger |x|.

Real-world note: damping and driving (what the ideal model ignores)

AP Physics 1 often emphasizes ideal SHM, but it’s worth knowing what breaks energy conservation:

• Damping (friction, air resistance, internal losses) removes mechanical energy, so amplitude decreases over time.
• A driving force (like pushing a swing) can add energy.

In a damped oscillator without a driver, E decreases, so the amplitude decreases because E=\frac{1}{2}kA^2 no longer stays constant.

You usually won’t need advanced damping mathematics in this section, but conceptually you should be ready to say: “Nonconservative forces change the total mechanical energy, so amplitude changes.”

Exam Focus

• Typical question patterns:
  • Use conservation of energy to find speed at a given displacement for a mass–spring oscillator, or find displacement given a speed.
  • Given amplitude (or maximum stretch) and k, find total energy and then find v_{\max}.
  • For pendulums, use mgh to relate a height change to speed at the bottom (often as a bridge between oscillations and energy).
• Common mistakes:
  • Confusing amplitude A with instantaneous displacement x when plugging into energy formulas; E uses A, while U_s uses x.
  • Forgetting that speed depends on position: using v_{\max} everywhere instead of computing v(x).
  • Treating mechanical energy as conserved even when the problem explicitly includes friction or energy loss; if damping is present, you cannot set initial total energy equal to final total energy without accounting for work done by nonconservative forces.