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Oscillations and Linearized Motion

13.1 Simple Harmonic Motion from Newton's Second Law

A Mass on a Spring

Hang a mass from a spring. Pull it down a little. Let go.

It bounces. Up and down, up and down, with a rhythm so steady you could set a watch by it. Not approximately steady --- remarkably steady. The mass does not speed up, does not slow down, does not drift. It traces the same path, at the same pace, over and over.

Why?

What is it about a spring that produces this relentless repetition? What determines how fast the mass bounces --- is it the mass, the spring, the distance you pull it, or all three? And why does this exact same behavior show up in systems that look nothing like a mass on a spring --- pendulum clocks, guitar strings, vibrating molecules, electrical circuits, even the wobble of a skyscraper in the wind?

[Video: A mass hangs from a spring, at rest. A hand reaches in and pulls the mass downward, then releases. The mass oscillates smoothly up and down. The camera holds steady as the oscillation repeats --- five, six, seven cycles, each one indistinguishable from the last. A timer in the corner counts the period. Then a cut: the same behavior in a pendulum clock, a tuning fork, a plucked guitar string, a ball rolling in a bowl. Each system looks different, but the rhythm of the back-and-forth motion is the same kind of repetition.]

This section is about understanding that rhythm. We will start with the simplest oscillating system --- a mass on a spring --- and discover the equation that governs its motion. The answer will turn out to be one of the most important equations in all of physics.

Before you read on: A mass on a spring oscillates up and down. Suppose you double the mass, keeping the same spring and pulling it down by the same amount.

Does the period (the time for one complete oscillation):

(a) double, (b) halve, (c) stay the same, or (d) change by some other factor?

Commit to your answer before reading on. Write it down.

Most people guess (a) --- double the mass, double the period. It feels right. A heavier mass should be more sluggish, so it should take longer. And it does take longer, but not twice as long. The period increases by a factor of $\sqrt{2} \approx 1.41$. Doubling the mass makes the oscillation about 41% slower, not 100% slower.

That is a strange result. Where does the square root come from? To answer that, we need to look at what the spring actually does to the mass as it moves.

Why Does It Repeat?

Let's trace the motion carefully, one step at a time.

The mass hangs at rest on the spring. The spring is stretched just enough to support the weight. This is the equilibrium position --- the place where the net force is zero and the mass would sit forever if left alone.

Now you pull the mass down a little below equilibrium and release it. What happens?

The spring is now stretched more than it needs to be. It pulls the mass upward, back toward equilibrium. The mass accelerates upward.

As the mass rises toward equilibrium, the spring gradually returns to its natural length. The restoring force shrinks. But the mass is now moving --- it has velocity. When it passes through equilibrium, the net force is zero, but the mass does not stop. It overshoots. It keeps moving upward.

Now the mass is above equilibrium. The spring is compressed (or not stretched enough). It pushes the mass back down. The mass decelerates, slows, stops momentarily at the top, and then begins falling back toward equilibrium.

The whole cycle repeats. Down through equilibrium, overshoot, slow down, stop at the bottom, reverse, back up through equilibrium, overshoot, slow down, stop at the top, reverse.

Pause and think: What makes this motion repeat? What is the essential ingredient?

Think about what would happen if the spring just pushed the mass with a constant force, regardless of position. Would the motion still repeat?

The essential ingredient is that the force always points toward equilibrium, and its strength is proportional to how far away you are. If you are displaced a lot, the force pulling you back is large. If you are displaced a little, the force is small. This is a restoring force: it always acts to restore the system to equilibrium, and it gets stronger the farther you stray.

A constant force would not produce oscillation. It would just push the mass in one direction forever. A restoring force that grows with displacement --- that is what creates the back-and-forth rhythm.

Exploring the Motion

Before we write any equations, spend time with the system. Adjust it. Watch what changes and what does not.

[Interactive: Mass-Spring Oscillator. A mass hangs from a vertical spring. The student can pull the mass down to any initial displacement and release it. The mass oscillates, and a real-time graph of $x(t)$ (position versus time) builds on the right side of the screen. Below the spring are three sliders:

  • Mass $m$: adjustable from 0.5 kg to 5 kg
  • Spring constant $k$: adjustable from 5 N/m to 50 N/m
  • Initial displacement $A$: adjustable from 1 cm to 15 cm

As each slider changes, the oscillation updates in real time. Numerical readouts display the current period $T$, frequency $f$, angular frequency $\omega$, and amplitude $A$. A small acceleration arrow on the mass shows the instantaneous acceleration, always pointing toward the equilibrium position.

Guided prompts appear one at a time:

  1. "Set $m = 1$ kg and $k = 10$ N/m. Pull the mass down 5 cm and release. Watch the graph of $x(t)$. What shape does it trace out?"
  2. "Now increase the mass to 4 kg, keeping everything else the same. What happens to the period? By what factor does it change?"
  3. "Reset to $m = 1$ kg. Now increase the spring constant from 10 to 40 N/m. What happens to the period?"
  4. "Now the key test: change the initial displacement --- the amplitude. Try 3 cm, then 6 cm, then 12 cm. What happens to the period?"
  5. "Watch the acceleration arrow as the mass moves. Where is the acceleration largest? Where is it zero? How does its direction relate to the position?"]

If you worked through those prompts carefully, you discovered three things:

  1. Increasing the mass makes the oscillation slower (longer period). But doubling the mass does not double the period --- it multiplies the period by about 1.41.
  2. Increasing the spring constant makes the oscillation faster (shorter period). A stiffer spring means a quicker bounce.
  3. Changing the amplitude does not change the period at all. A mass pulled down 3 cm and a mass pulled down 12 cm take the same time to complete one cycle.

That third discovery is the most surprising. Pull the mass farther, and it has more ground to cover. You would expect it to take longer. But it also experiences a stronger restoring force when it is farther from equilibrium, so it moves faster. These two effects --- more distance, more speed --- cancel exactly, and the period stays the same.

This property has a name: isochronism. The period of a mass-spring oscillation is independent of amplitude. It is one of the defining features of the motion we are about to analyze.

From Force to Equation

Now let's build the mathematics. We have a mass $m$ attached to a spring with spring constant $k$. We measure position $x$ from the equilibrium point, so $x = 0$ is where the mass sits when undisturbed.

The spring obeys Hooke's law: the force it exerts on the mass is proportional to the displacement and directed toward equilibrium:

$$F = -kx$$

The minus sign is crucial. When $x$ is positive (mass displaced to the right or above equilibrium), the force is negative (pointing back toward equilibrium). When $x$ is negative (mass displaced to the left or below equilibrium), the force is positive (again pointing back toward equilibrium). The force always opposes the displacement. That is what makes it restoring.

Now apply Newton's second law: the net force on the mass equals mass times acceleration.

$$ma = -kx$$

Since acceleration is the second derivative of position with respect to time, $a = \frac{d^2x}{dt^2}$, this becomes:

$$m\frac{d^2x}{dt^2} = -kx$$

Divide both sides by $m$:

$$\frac{d^2x}{dt^2} = -\frac{k}{m}x$$

Stop here for a moment and read that equation. It says: the acceleration of the mass is proportional to its position, but in the opposite direction. Farther from equilibrium means stronger acceleration back toward it. Closer to equilibrium means weaker acceleration. At equilibrium itself, the acceleration is zero --- but by then the mass is already moving.

This equation appeared before. In Section 4.3, you encountered the acceleration model $a = -\omega^2 x$ and observed that it produces oscillation. At that point, you saw the behavior but did not solve the equation. Now we solve it.

Solving the Equation

The equation

$$\frac{d^2x}{dt^2} = -\frac{k}{m}x$$

is a second-order linear differential equation with constant coefficients. You do not need that full name to understand it. What matters is the question it asks: what function $x(t)$, when you take its second derivative, gives back the same function multiplied by a negative constant?

Think about that for a moment. You need a function whose second derivative is a negative multiple of itself.

Exponential functions do not work: $\frac{d^2}{dt^2}(e^{bt}) = b^2 e^{bt}$, which is a positive multiple. Polynomials do not work: $\frac{d^2}{dt^2}(t^2) = 2$, which is not proportional to $t^2$. Linear functions do not work either.

But cosine does. Take $x(t) = \cos(\omega t)$:

$$\frac{dx}{dt} = -\omega \sin(\omega t)$$

$$\frac{d^2x}{dt^2} = -\omega^2 \cos(\omega t) = -\omega^2 \, x(t)$$

The second derivative gives back the original function times $-\omega^2$. That is exactly what we need, as long as

$$\omega^2 = \frac{k}{m} \qquad \Longrightarrow \qquad \omega = \sqrt{\frac{k}{m}}$$

Sine works too: $\frac{d^2}{dt^2}[\sin(\omega t)] = -\omega^2 \sin(\omega t)$. And any combination of sine and cosine works. The most general solution is:

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

where:

  • $A$ is the amplitude --- the maximum displacement from equilibrium.
  • $\omega = \sqrt{k/m}$ is the angular frequency --- how rapidly the oscillation cycles, measured in radians per second.
  • $\phi$ is the phase constant --- it encodes where in the cycle the motion starts at $t = 0$.

This is simple harmonic motion (SHM): sinusoidal oscillation at a frequency determined entirely by the system parameters $k$ and $m$.

What the Solution Tells Us

Let's unpack what we just found.

The angular frequency depends only on $k$ and $m$.

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

A stiffer spring (larger $k$) means faster oscillation. A heavier mass (larger $m$) means slower oscillation. The amplitude $A$ does not appear anywhere in this expression. This is the mathematical reason for isochronism: the frequency is set by the system, not by how hard you pull.

The period is

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

Now we can answer the opening prediction. If you double the mass, the new period is:

$$T_{\text{new}} = 2\pi\sqrt{\frac{2m}{k}} = \sqrt{2} \cdot 2\pi\sqrt{\frac{m}{k}} = \sqrt{2} \, T$$

The period increases by a factor of $\sqrt{2}$, not by a factor of 2. The square root relationship between mass and period is a direct consequence of the equation of motion.

The frequency is

$$f = \frac{1}{T} = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$$

This is measured in cycles per second (hertz, Hz). The angular frequency $\omega$ is measured in radians per second. They are related by $\omega = 2\pi f$.

The amplitude and phase come from initial conditions.

The equation of motion determines $\omega$. It does not determine $A$ or $\phi$. Those are set by how you start the system --- how far you pull the mass and how fast it is moving when you let go. This is exactly the initial-value problem framework from Chapter 4: the differential equation gives the family of solutions, and the initial conditions pick out the specific one.

For example, if you pull the mass to a displacement $x_0$ and release it from rest:

  • At $t = 0$: $x(0) = A\cos(\phi) = x_0$
  • At $t = 0$: $v(0) = -A\omega\sin(\phi) = 0$

The second equation gives $\sin(\phi) = 0$, so $\phi = 0$. The first then gives $A = x_0$. The solution is:

$$x(t) = x_0 \cos(\omega t)$$

Simple, clean, and completely determined.

Why Sinusoidal? A Deeper Look

You might accept that cosine is a solution without quite understanding why the motion must be sinusoidal. Here is a way to see it that goes beyond "plug it in and check."

The equation $a = -\omega^2 x$ creates a feedback loop:

  1. The mass is displaced from equilibrium.
  2. The restoring force accelerates it back toward equilibrium.
  3. The mass picks up speed and overshoots equilibrium.
  4. Now displaced on the other side, the restoring force reverses and accelerates it back again.
  5. It overshoots again. The cycle repeats.

At each moment, the acceleration is recalculated from the current position. The force is never "too strong" or "too weak" in a way that would cause the oscillation to grow or shrink --- it is always exactly proportional to the displacement. This proportionality is what keeps the motion perfectly periodic and perfectly sinusoidal.

If the force were stronger than proportional at large displacements, the mass would come back faster and the oscillation would not be a pure sinusoid. If the force were weaker than proportional, the mass would come back slower. Only the linear restoring force $F = -kx$ produces the exact sinusoidal motion we call simple harmonic motion.

The linearity of Hooke's law is what makes the oscillation simple. Change the force law, and you still get oscillation --- but it is no longer sinusoidal, and no longer "simple" harmonic.

A Historical Note

In 1656, the Dutch mathematician and physicist Christiaan Huygens built the first pendulum clock. His key insight was isochronism: for small swings, the period of a pendulum does not depend on how wide the swing is. A large swing and a small swing take the same time.

Before Huygens, clocks were notoriously unreliable. Mechanical clocks with balance wheels could drift by fifteen minutes a day. Huygens' pendulum clock was accurate to about ten seconds per day --- an improvement by a factor of nearly a hundred. The entire history of precision timekeeping, from Huygens' pendulum to the quartz oscillators in modern watches, rests on the same mathematical fact you just derived: the period of a simple harmonic oscillator is independent of its amplitude.

Isochronism is not an accident. It is a consequence of the equation $\frac{d^2x}{dt^2} = -\omega^2 x$. The amplitude $A$ appears in the solution $x(t) = A\cos(\omega t + \phi)$ but not in the angular frequency $\omega = \sqrt{k/m}$. No matter how large or small the oscillation, the period is the same. That mathematical structure made accurate clocks possible --- and it shows up in every oscillating system governed by a linear restoring force.

Connecting to What You Already Know

This result did not appear out of nowhere. You have seen the pieces before.

In Section 4.3, you studied the acceleration model $a = -\omega^2 x$ and observed that it produces oscillatory behavior. You saw the motion in graphs and simulations, and you noted that it was qualitatively different from constant-acceleration or velocity-dependent models. But you did not solve the differential equation. Now you have: the solution is $x(t) = A\cos(\omega t + \phi)$.

In Section 7.5, you used energy diagrams to analyze motion near potential energy minima. Near a minimum, the potential energy curve is approximately parabolic: $U(x) \approx \frac{1}{2}kx^2$. The force derived from that potential is $F = -dU/dx = -kx$ --- exactly Hooke's law. So the oscillation of a mass on a spring is the same motion you saw qualitatively in energy diagrams. Later in this chapter, we will return to energy methods and make this connection precise.

In Chapter 5, you applied Newton's second law $\sum \vec{F} = m\vec{a}$ to analyze the motion of objects under various forces. Here, the force is $F = -kx$, and the resulting equation of motion $\frac{d^2x}{dt^2} = -\frac{k}{m}x$ is just Newton's second law for a specific force model. The machinery is the same --- only the force law is new.

Practice

Layer 1: Concrete

A block of mass $m = 0.50$ kg is attached to a spring with spring constant $k = 200$ N/m. The block is pulled 4.0 cm from equilibrium and released from rest.

Find: (a) the angular frequency $\omega$, (b) the period $T$, (c) the frequency $f$, and (d) write the position as a function of time $x(t)$.

Check your answer (a) The angular frequency is $$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.50}} = \sqrt{400} = 20 \text{ rad/s}$$ (b) The period is $$T = \frac{2\pi}{\omega} = \frac{2\pi}{20} = 0.314 \text{ s}$$ The block completes about 3.2 oscillations every second. (c) The frequency is $$f = \frac{1}{T} = \frac{1}{0.314} = 3.18 \text{ Hz}$$ Or equivalently, $f = \omega / (2\pi) = 20/(2\pi) = 3.18$ Hz. (d) The block is released from rest at $x = 4.0$ cm, so $A = 0.040$ m and $\phi = 0$: $$x(t) = 0.040 \cos(20t) \text{ m}$$ where $t$ is in seconds and $x$ in meters.

Layer 2: Pattern

For each change below, predict whether the period of a mass-spring oscillator increases, decreases, or stays the same. Then explain your reasoning using $T = 2\pi\sqrt{m/k}$.

(a) The mass is quadrupled (keeping $k$ the same).

(b) The spring constant is quadrupled (keeping $m$ the same).

(c) The amplitude is doubled (keeping $m$ and $k$ the same).

(d) Both $m$ and $k$ are doubled simultaneously.

Check your answer (a) **Period doubles.** $T = 2\pi\sqrt{m/k}$. Replace $m$ with $4m$: $T_{\text{new}} = 2\pi\sqrt{4m/k} = 2 \cdot 2\pi\sqrt{m/k} = 2T$. Quadrupling the mass doubles the period because $T \propto \sqrt{m}$. (b) **Period halves.** Replace $k$ with $4k$: $T_{\text{new}} = 2\pi\sqrt{m/(4k)} = \frac{1}{2} \cdot 2\pi\sqrt{m/k} = T/2$. A stiffer spring produces a faster oscillation. (c) **Period stays the same.** The amplitude $A$ does not appear in the formula $T = 2\pi\sqrt{m/k}$. This is isochronism --- the period is independent of amplitude. (d) **Period stays the same.** $T_{\text{new}} = 2\pi\sqrt{2m/(2k)} = 2\pi\sqrt{m/k} = T$. Doubling both $m$ and $k$ leaves the ratio $m/k$ unchanged, so the period does not change. Notice the pattern: only the *ratio* $m/k$ matters for the period. Any change that preserves this ratio preserves the period.

Layer 3: Structure

Why is the period of a mass-spring oscillator independent of amplitude? What feature of the equation of motion $\frac{d^2x}{dt^2} = -\frac{k}{m}x$ guarantees this?

Check your answer The key feature is **linearity**: the restoring force is directly proportional to the displacement ($F = -kx$). This means that when you double the displacement, the force doubles as well. Consider what happens when you double the amplitude. The mass now has to travel twice as far in each half-cycle. But at every point in its motion, the force (and therefore the acceleration) is also twice as large. Greater acceleration means the mass moves faster. The extra distance and the extra speed cancel exactly, and the time for one cycle remains the same. Mathematically, this shows up because the equation $\frac{d^2x}{dt^2} = -\omega^2 x$ is linear in $x$. If $x(t)$ is a solution, then $cx(t)$ is also a solution for any constant $c$ --- and it has the same $\omega$. Scaling the amplitude scales the entire solution vertically without stretching or compressing it in time. The angular frequency $\omega = \sqrt{k/m}$ depends only on the system parameters, not on the amplitude. If the force were nonlinear --- say $F = -kx^3$ --- doubling the displacement would *more* than double the force, and the cancellation would not be exact. The period would depend on amplitude, and the motion would no longer be sinusoidal.

Layer 4: Debug

A student is analyzing a mass on a spring. They reason as follows:

"If I double the amplitude, the mass has to travel twice as far. Distance equals speed times time, and the speed doesn't change. So the period must double."

This reasoning leads to the prediction that doubling the amplitude doubles the period. But we know the period is independent of amplitude. Find the error in the student's reasoning and explain what actually happens.

Check your answer The error is in the claim that "the speed doesn't change." When you double the amplitude, the maximum speed *does* change --- it doubles. Here is why. The velocity in SHM is $v(t) = -A\omega\sin(\omega t + \phi)$. The maximum speed is $v_{\max} = A\omega$. Doubling $A$ doubles $v_{\max}$. So the student's argument has two pieces: - Distance doubles. **Correct.** - Speed stays the same. **Incorrect --- speed also doubles.** Since both the total distance traveled and the characteristic speed increase by the same factor, the time to complete one cycle stays the same. The student's error was treating the speed as fixed when it actually scales with amplitude. Physically, this makes sense. A larger displacement means a stronger restoring force ($F = -kx$), which means greater acceleration, which means the mass moves faster. The proportionality between force and displacement is exactly what guarantees this perfect cancellation. The time comes out the same regardless of how far you pull.

Reflection

Think back over what you have explored in this section.

What makes a restoring force "restoring" --- and why does it lead to oscillation rather than some other kind of motion?

Consider this: a constant force produces constant acceleration (the motion of a thrown ball). A velocity-dependent force can produce exponential decay (a parachutist approaching terminal velocity). A force proportional to displacement but pointing away from equilibrium produces exponential growth (an unstable system that flies apart). But a force proportional to displacement and pointing toward equilibrium produces oscillation --- repeating, endlessly cycling motion.

What is special about the combination of "proportional to displacement" and "directed toward equilibrium"? Why does this combination, and only this combination, produce the clean sinusoidal rhythm of simple harmonic motion?

Looking Ahead

You have just derived the most fundamental equation of oscillatory physics: $\frac{d^2x}{dt^2} = -\omega^2 x$, with solution $x(t) = A\cos(\omega t + \phi)$. The frequency is set by the system, the amplitude and phase are set by how you start it, and --- remarkably --- the period does not depend on how far you pull.

But we analyzed this motion entirely through forces: $F = ma$, with $F = -kx$. There is another way to understand oscillation --- through energy. A mass on a spring is constantly trading kinetic energy for potential energy and back again, with the total energy staying fixed. This energy viewpoint will give you a powerful second tool for analyzing oscillations, one that connects directly to the energy diagrams you studied in Section 7.5.

In the next section, we explore this energy perspective. You will see that the oscillation can be understood as energy sloshing between two reservoirs --- kinetic and potential --- and that a single energy equation can answer questions about speed and position without solving the differential equation at all.