13.3 Angular Frequency, Phase, and Amplitude

The Same Motion, But Different

[Video: Two identical pendulums hang from the same rod. Someone pulls the first one back and releases it. It swings smoothly, back and forth. A few seconds later, someone pulls the second pendulum back by the same distance and releases it. Both pendulums swing at the same rate, sweeping through the same arc --- but they are never in the same place at the same time. They are permanently out of step.]

Watch those two pendulums. They have the same length, so they swing at the same rate. They were pulled back by the same distance, so they swing through the same arc. And yet, if you watch them side by side, they are doing something observably different. One reaches the top of its swing while the other is passing through the bottom.

What is different about them?

It is not how fast they oscillate --- that is identical. It is not how far they swing --- that is identical too. The difference is when they started. The first pendulum had a head start. It is further along in its cycle at any given moment.

There must be a parameter that captures this difference. And there must be a way to write it down in the mathematics. That parameter is called the phase, and it is one of three numbers that completely characterize a sinusoidal oscillation.

Before you read on: Two oscillators have the same angular frequency $\omega$ and the same amplitude $A$, but different phases $\phi$. At $t = 0$, one oscillator is at its maximum displacement (all the way to the right). The other is at the equilibrium position (the center), moving in the negative direction.

Using $x(t) = A\cos(\omega t + \phi)$, what are the phase values $\phi$ for each oscillator?

Write down your answers before continuing.

Three Parameters, Three Roles

In Sections 13.1 and 13.2, you saw that the solution to the simple harmonic oscillator equation $\ddot{x} = -\omega^2 x$ is

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

This expression contains exactly three parameters: $A$, $\omega$, and $\phi$. Each one controls a different aspect of the motion. But what exactly does each one do? And where does each one come from?

The best way to answer these questions is to see the effects directly.

[Interactive: Parameter Explorer. A single oscillator is displayed alongside its $x(t)$ graph. Three sliders control amplitude $A$, angular frequency $\omega$, and phase $\phi$. The $x(t)$ curve updates in real time as each slider is adjusted.

Instructions guide the student through a structured exploration:

"Set $\omega = 2$ and $\phi = 0$. Now move the amplitude slider from $A = 1$ to $A = 3$. What changes about the graph? What stays the same?"
"Reset to $A = 2$, $\phi = 0$. Now increase $\omega$ from 1 to 3. What changes? What stays the same?"
"Reset to $A = 2$, $\omega = 2$. Now change $\phi$ from $0$ to $\pi/2$ to $\pi$ to $3\pi/2$. What happens to the graph? Describe the change in your own words."

A second panel shows the oscillator as a physical animation --- a mass sliding on a frictionless surface attached to a spring --- so students can connect the graph to the motion.]

Pause and think: Before reading the next section, describe in your own words what each parameter does to the graph. Which parameter changes the height of the curve? Which changes how quickly it repeats? Which shifts it left or right?

Amplitude: How Far

The parameter $A$ is the amplitude of the oscillation. It tells you the maximum displacement from equilibrium --- the farthest the object gets from the center.

When you change $A$, the graph stretches vertically. The peaks get higher and the troughs get deeper, but the timing of the oscillation does not change. The peaks still occur at the same moments. The zero crossings stay put.

Physically, amplitude corresponds to how much energy you put in. A larger pull gives a larger amplitude. The system oscillates more vigorously, but at the same rate.

$$x_{\max} = A, \qquad x_{\min} = -A.$$

One subtle point: amplitude is always a positive number. It represents a distance, not a direction. Writing $A = -3$ does not mean the oscillator swings "less far" --- it means you have folded a sign into the amplitude that properly belongs in the phase. In this course, we take $A > 0$ by convention, and let $\phi$ handle the sign.

Angular Frequency: How Fast

The parameter $\omega$ is the angular frequency. It controls how rapidly the oscillation repeats.

When you increase $\omega$, the oscillation cycles faster. The peaks crowd together. The graph compresses horizontally. But the height of the peaks does not change --- the system swings just as far, just more quickly.

Angular frequency is measured in radians per second. The word "angular" is there because of the deep connection between oscillatory motion and circular motion (which you will explore in Section 13.6). One full cycle of the cosine function corresponds to $2\pi$ radians of the argument, so the time for one complete cycle is

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

This is the period --- the time for one complete oscillation. The frequency $f$ is the number of complete cycles per second:

$$f = \frac{1}{T} = \frac{\omega}{2\pi}.$$

Frequency is measured in hertz (Hz), where 1 Hz means one cycle per second. Angular frequency and ordinary frequency carry the same information, just packaged differently. You can convert freely between them:

$$\omega = 2\pi f = \frac{2\pi}{T}.$$

Here is the crucial fact about $\omega$: it is determined by the system, not by how you start the motion. For a mass on a spring, $\omega = \sqrt{k/m}$. Change the spring constant or the mass, and $\omega$ changes. But pull the mass farther back, or start it with a push, and $\omega$ stays exactly the same. The rate of oscillation is built into the system itself.

This is worth saying again, because it surprises many students. If you double the amplitude, the period does not change. If you start the oscillation from a different position, the period does not change. The angular frequency $\omega$ is a property of the system (the spring and the mass), not a property of how you launch the oscillation.

Phase: Where in the Cycle

The parameter $\phi$ is the phase (or more precisely, the initial phase or phase constant). It tells you where in the cycle the oscillator is at $t = 0$.

When you change $\phi$, the graph shifts horizontally. The shape of the curve --- its height, its spacing between peaks --- stays the same. But the curve slides left or right along the time axis. A different phase means the oscillation starts at a different point in its cycle.

To see why, look at the argument of the cosine:

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

At $t = 0$, the position is

$$x(0) = A\cos(\phi).$$

So $\phi$ determines where the oscillator begins:

$\phi = 0$: the oscillator starts at maximum positive displacement, $x(0) = A$. It begins at the top of its swing and initially moves toward equilibrium.
$\phi = \pi/2$: the oscillator starts at equilibrium, $x(0) = 0$. Taking the derivative, $v(0) = -A\omega\sin(\pi/2) = -A\omega$. It begins at the center, moving in the negative direction.
$\phi = \pi$: the oscillator starts at maximum negative displacement, $x(0) = -A$.
$\phi = 3\pi/2$: the oscillator starts at equilibrium again, $x(0) = 0$, but now $v(0) = -A\omega\sin(3\pi/2) = A\omega$. It begins at the center, moving in the positive direction.

Check your prediction: Go back to your prediction from the beginning of this section. The oscillator at maximum displacement has $\phi = 0$. The oscillator at equilibrium, moving in the negative direction, has $\phi = \pi/2$. Did you get these right?

If you wrote $\phi = \pi/2$ for the second one, good --- the negative sine at $\pi/2$ gives negative initial velocity. If you wrote $\phi = -\pi/2$ or $\phi = 3\pi/2$, check the sign of the velocity. These correspond to the oscillator starting at equilibrium but moving in the positive direction.

The phase is sometimes the hardest of the three parameters for students to visualize, because it does not correspond to a simple measurement like "how far" or "how fast." It is more like a clock hand: it tells you where the oscillator is in its cycle at a chosen reference time. Two oscillators with the same $A$ and $\omega$ but different $\phi$ are doing exactly the same motion, just offset in time.

The Full Argument: What Is $\omega t + \phi$?

The quantity $\omega t + \phi$ inside the cosine is called the phase of the oscillation at time $t$ (as opposed to $\phi$ alone, which is the phase at $t = 0$). The full phase grows linearly in time:

$$\text{phase at time } t = \omega t + \phi.$$

Every time the phase increases by $2\pi$, the cosine completes one full cycle and the motion repeats. This is another way to see why the period is $T = 2\pi/\omega$: in one period, the phase must advance by exactly $2\pi$.

$$\omega T = 2\pi \implies T = \frac{2\pi}{\omega}.$$

Where the Parameters Come From

Now here is the question that ties everything together: where do $A$, $\omega$, and $\phi$ actually come from?

The answer splits neatly into two categories:

Parameter	Determined by	Example
$\omega$	The system itself	$\omega = \sqrt{k/m}$ for a mass-spring system
$A$	Initial conditions	How far you pull the mass before releasing it
$\phi$	Initial conditions	Where the mass is, and how fast it is moving, at $t = 0$

This is exactly the structure of an initial-value problem, which you studied in Section 4.2. The differential equation $\ddot{x} = -\omega^2 x$ defines a family of solutions --- all possible sinusoidal motions with angular frequency $\omega$. The initial conditions $x(0)$ and $v(0)$ select a specific member of that family by determining $A$ and $\phi$.

Let's make this explicit. If you know that $x(0) = x_0$ and $v(0) = v_0$, then from the solution $x(t) = A\cos(\omega t + \phi)$ and its derivative $v(t) = -A\omega\sin(\omega t + \phi)$:

$$x_0 = A\cos\phi, \qquad v_0 = -A\omega\sin\phi.$$

You can solve these two equations for $A$ and $\phi$:

$$A = \sqrt{x_0^2 + \left(\frac{v_0}{\omega}\right)^2},$$

$$\tan\phi = -\frac{v_0}{\omega x_0}.$$

(You need to use the signs of $\cos\phi$ and $\sin\phi$ individually to place $\phi$ in the correct quadrant, since the tangent function repeats every $\pi$.)

The amplitude depends on both the initial position and the initial velocity. This makes physical sense: you can create a large-amplitude oscillation either by pulling the mass far from equilibrium (large $x_0$) or by giving it a strong push at the center (large $v_0$).

Variation: Isolating Each Parameter

One of the most powerful ways to build understanding is to change one thing at a time and watch what happens. Let's do this systematically.

[Interactive: Variation Explorer. Three oscillators are displayed simultaneously, each with its own $x(t)$ graph. The student is guided through three rounds of comparison:

Round 1: Vary $\phi$, hold $A$ and $\omega$ fixed. Three oscillators all have $A = 2$, $\omega = 3$, but $\phi = 0$, $\phi = \pi/3$, and $\phi = 2\pi/3$. The graphs are superimposed in different colors. - Prompt: "What is the same about all three curves? What is different? Describe what $\phi$ does in one sentence."

Round 2: Vary $A$, hold $\omega$ and $\phi$ fixed. Three oscillators all have $\omega = 3$, $\phi = 0$, but $A = 1$, $A = 2$, and $A = 3$. Graphs are superimposed. - Prompt: "What changed? What stayed the same? Where do all three curves cross zero at the same time?"

Round 3: Vary $\omega$, hold $A$ and $\phi$ fixed. Three oscillators all have $A = 2$, $\phi = 0$, but $\omega = 1$, $\omega = 2$, and $\omega = 3$. Graphs are superimposed. - Prompt: "What changed? What stayed the same? Do they all start at the same position? Do they return to the start at the same time?"]

These three comparisons isolate each parameter cleanly:

Varying $\phi$ shifts the graphs horizontally. The shape is identical; only the starting point changes. All three curves have the same amplitude and the same period.
Varying $A$ stretches the graphs vertically. All three curves hit zero at exactly the same times. The peaks and troughs grow taller, but the timing is unchanged.
Varying $\omega$ compresses or stretches the graphs horizontally. The first peak comes sooner when $\omega$ is larger. The period shrinks. But all three start at the same height (because $\phi = 0$ and $x(0) = A$ in each case).

Notice the contrast between varying $\phi$ and varying $\omega$. Both affect the horizontal appearance of the graph, but in different ways. Changing $\phi$ slides the whole curve left or right without altering the spacing between peaks. Changing $\omega$ compresses or stretches the spacing itself. The difference is between shifting a pattern and changing the pattern.

Reading Parameters from Graphs

In practice, you will often need to extract $A$, $\omega$ (or $T$), and $\phi$ from a graph of $x(t)$. Here is how each parameter shows up visually:

Amplitude $A$: The distance from the horizontal axis (equilibrium) to the highest peak. Equivalently, half the distance from the highest peak to the lowest trough.
Period $T$: The time between two successive peaks (or two successive troughs, or any two points where the motion repeats). Then $\omega = 2\pi / T$.
Phase $\phi$: Look at where the curve is at $t = 0$. If it starts at its maximum, $\phi = 0$. If it starts at zero and heading downward, $\phi = \pi/2$. In general, $\phi$ is determined by $x(0) = A\cos\phi$ together with the sign of $v(0)$.

Reading the phase from a graph takes more care than reading $A$ or $T$. You cannot determine $\phi$ from the position at $t = 0$ alone --- you also need to know the direction of motion (whether the velocity is positive or negative at $t = 0$). A curve that passes through $x = 0$ at $t = 0$ might have $\phi = \pi/2$ or $\phi = 3\pi/2$, depending on which way the oscillator is moving.

A Summary Table

Symbol	Name	What it controls	Determined by	Units
$A$	Amplitude	How far from equilibrium	Initial conditions $(x_0, v_0)$	meters (same as $x$)
$\omega$	Angular frequency	How fast the oscillation repeats	The system $(k/m)$	rad/s
$\phi$	Phase (initial phase)	Where in the cycle at $t = 0$	Initial conditions $(x_0, v_0)$	radians (dimensionless)
$T = 2\pi/\omega$	Period	Time for one full cycle	The system	seconds
$f = 1/T$	Frequency	Cycles per second	The system	Hz (s$^{-1}$)

Practice

Layer 1: Concrete

A $0.5\,\text{kg}$ mass is attached to a spring with $k = 8\,\text{N/m}$. At $t = 0$, the mass is displaced $0.1\,\text{m}$ to the right of equilibrium and released from rest.

Find the angular frequency $\omega$, the amplitude $A$, and the phase $\phi$. Then write the complete solution $x(t)$.

Check your answer

The angular frequency comes from the system: $$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{8}{0.5}} = \sqrt{16} = 4\,\text{rad/s}.$$ The initial conditions are $x(0) = 0.1\,\text{m}$ and $v(0) = 0$ (released from rest). From $x(0) = A\cos\phi = 0.1$ and $v(0) = -A\omega\sin\phi = 0$: Since $v(0) = 0$ and $A\omega \neq 0$, we need $\sin\phi = 0$, so $\phi = 0$ or $\phi = \pi$. Since $x(0) = A\cos\phi = 0.1 > 0$ and $A > 0$, we need $\cos\phi > 0$, so $\phi = 0$. Then $A = 0.1\,\text{m}$. The complete solution is: $$x(t) = 0.1\cos(4t) \quad \text{(in meters, with } t \text{ in seconds)}.$$ The period is $T = 2\pi/4 = \pi/2 \approx 1.57\,\text{s}$ and the frequency is $f = 1/T \approx 0.64\,\text{Hz}$.

Layer 2: Pattern

Three $x(t)$ graphs are shown below. All represent simple harmonic motion. Match each graph to the correct parameter set.

Graph A: Starts at $x = 0$ at $t = 0$, first moves in the positive direction, reaches $x = 3$ at $t = \pi/4$.

Graph B: Starts at $x = 3$ at $t = 0$, first peak at $t = 0$, period is $\pi$.

Graph C: Starts at $x = -3$ at $t = 0$, period is $\pi$.

Parameter sets: 1. $A = 3$, $\omega = 2$, $\phi = 0$ 2. $A = 3$, $\omega = 2$, $\phi = \pi$ 3. $A = 3$, $\omega = 2$, $\phi = -\pi/2$

Check your answer

**Graph B matches Parameter Set 1** ($A = 3$, $\omega = 2$, $\phi = 0$). Check: $x(0) = 3\cos(0) = 3$. The curve starts at its maximum, so $t = 0$ is a peak. Period is $T = 2\pi/\omega = 2\pi/2 = \pi$. This all fits Graph B. **Graph C matches Parameter Set 2** ($A = 3$, $\omega = 2$, $\phi = \pi$). Check: $x(0) = 3\cos(\pi) = -3$. The curve starts at maximum negative displacement. Period is still $T = \pi$. This fits Graph C. **Graph A matches Parameter Set 3** ($A = 3$, $\omega = 2$, $\phi = -\pi/2$). Check: $x(0) = 3\cos(-\pi/2) = 0$. The velocity at $t = 0$ is $v(0) = -3(2)\sin(-\pi/2) = 6 > 0$, so the oscillator starts at zero and moves in the positive direction. At $t = \pi/4$: $x(\pi/4) = 3\cos(2 \cdot \pi/4 - \pi/2) = 3\cos(0) = 3$. This matches Graph A. The key insight: all three graphs have the same amplitude and period. Only the phase differs, and it shows up as a horizontal shift of the curve.

Layer 3: Structure

Changing the phase $\phi$ shifts the $x(t)$ graph horizontally (left or right along the time axis), not vertically (up or down).

Why? Explain this using the mathematical structure of $x(t) = A\cos(\omega t + \phi)$. What would have to be different about the equation for changing a parameter to shift the graph vertically?

Check your answer

Look at where $\phi$ appears in the expression $A\cos(\omega t + \phi)$. It is *inside* the cosine, added to $\omega t$. This means $\phi$ modifies the *input* to the cosine function. Adding a constant to the input of any function shifts its graph horizontally. This is a general fact from function transformations: replacing $t$ with $t + c$ in $f(t)$ shifts the graph of $f$ to the left by $c$. Here, $x(t) = A\cos(\omega t + \phi) = A\cos\!\big(\omega(t + \phi/\omega)\big)$, which is the graph of $A\cos(\omega t)$ shifted to the left by $\phi/\omega$ seconds. To shift the graph *vertically*, you would need to add a constant *outside* the cosine: $$x(t) = A\cos(\omega t + \phi) + C.$$ This $C$ would represent a new equilibrium position --- the oscillation would be centered at $x = C$ instead of $x = 0$. In the pure SHM solution, there is no such term, because the equilibrium is at the origin by construction. So the reason changing $\phi$ shifts horizontally is structural: $\phi$ lives inside the argument of the cosine, where it acts as a time offset, not a position offset.

Layer 4: Debug

A student looks at an $x(t)$ graph and reads the period as "the time between two peaks." The student measures from the first peak to the second peak and gets $T = 1.5\,\text{s}$.

Is this a reliable method? Under what circumstances might it fail or be inaccurate?

Check your answer

The method is **correct in principle** --- the time between two consecutive peaks is indeed one period. The cosine function repeats every $T = 2\pi/\omega$, and consecutive maxima are separated by exactly one period. However, there are practical issues that can make this unreliable: 1. **Graph resolution.** Peaks are flat. The cosine curve has zero slope at its maximum, so the exact location of the peak is hard to pinpoint visually. A small error in reading the peak positions gets amplified into an error in $T$. Reading zero crossings is often more accurate, because the curve passes through zero with the steepest slope, making the crossing point sharper. 2. **Measuring over one period.** The student measured between just two peaks, giving a single period. A better approach is to measure over *many* periods (say, from the first peak to the eleventh peak) and divide by the number of intervals (ten). This averages out reading errors. 3. **The method assumes pure SHM.** If the oscillation is damped, the peaks are not evenly spaced in general (though for underdamped oscillations, the spacing is still $T$ --- only the amplitude decays). If the motion is not purely sinusoidal, the time between peaks may not equal the fundamental period. So the method is valid in theory but can be imprecise in practice. A more robust method is to count many cycles and divide, or to use zero crossings where the curve is steepest.

Reflection

Think about the three parameters $A$, $\omega$, and $\phi$.

Which of the three is determined by the system itself, and which are determined by how you start the motion?

This is a deep structural distinction. The angular frequency $\omega$ is fixed once you specify the spring and the mass --- no matter how you release the object, it oscillates at the same rate. The amplitude $A$ and phase $\phi$ are up to you. They encode the initial conditions --- the choices you made when you set the oscillation in motion.

You might also consider: is there a real-world system where you would want to control the phase of an oscillation precisely? (Think about music, radio signals, or the synchronized pendulums that started this section.) Why would the phase matter?

Looking Ahead

You now have the complete toolkit for describing simple harmonic motion: the angular frequency $\omega$ tells you how fast, the amplitude $A$ tells you how far, and the phase $\phi$ tells you where the cycle begins. Together with the equation $x(t) = A\cos(\omega t + \phi)$, these three numbers capture everything about a sinusoidal oscillation.

But so far, we have worked with systems where the restoring force is exactly $F = -kx$ --- a perfect linear spring. What about a pendulum? The restoring force for a pendulum involves $\sin\theta$, not $\theta$. In the next section, you will see how the small-angle approximation $\sin\theta \approx \theta$ turns the pendulum equation into the SHM equation, and you will learn how to judge when that approximation is good enough --- and when it breaks down.

The idea that complex equations can be simplified near equilibrium is one of the most powerful tools in physics. It is why SHM is everywhere.