Introduction: Spinning Round and Round, Moving Back and Forth
Hello, everyone! In this chapter, we will learn about "Circular Motion and Simple Harmonic Motion," which is one of the most frequently tested topics in physics on the Common Test.
You might think, "There are so many formulas, it sounds difficult..." but don't worry! In fact, once you see that simple harmonic motion is just the shadow (projection) of circular motion, it will make surprising sense.
Think of familiar phenomena in everyday life—like Ferris wheels and cars turning around curves (circular motion), or clock pendulums and vibrating guitar strings (simple harmonic motion)—as we master the basics step by step!
1. Uniform Circular Motion (Spinning Round and Round)
When an object moves along a circular path at a constant speed, it is called uniform circular motion. Even though we say "uniform speed," the direction of travel is constantly changing, which means the velocity (a vector) changes from moment to moment. In other words, uniform circular motion is an accelerated motion.
(1) Fundamental Physical Quantities (Angular Velocity, Period, Frequency)
First, let's organize the vocabulary used to describe circular motion.
・Angular velocity \( \omega \) (omega): The angle (central angle) through which the object rotates relative to the center of the circle in 1 second. The unit used is \( \text{rad/s} \) (radians per second).
・Period \( T \): The time it takes for the object to complete exactly one revolution (\( 2\pi \, \text{rad} \)) along the circular path. The unit is \( \text{s} \) (seconds).
・Frequency \( n \) (or \( f \)): The number of revolutions per second. The unit used is \( \text{Hz} \) (hertz).
・Speed \( v \): If the radius is \( r \), this is the instantaneous speed moving along the circumference. The unit is \( \text{m/s} \).
A very important and elegant set of relationships holds among these quantities:
\( v = r\omega \)
\( T = \frac{2\pi r}{v} = \frac{2\pi}{\omega} \)
\( n = \frac{1}{T} = \frac{\omega}{2\pi} \)
Key Point: As the radius \( r \) increases, even at the same angular velocity \( \omega \) (the same rotational pace), the speed \( v \) of an object on the outside increases. This is why horses on the outer edge of a carousel feel faster!
(2) Centripetal Acceleration and Centripetal Force
In circular motion, even though the speed is constant, the direction continuously changes, which produces a centripetal acceleration directed toward the center.
Using radius \( r \), angular velocity \( \omega \), and speed \( v \), the magnitude of centripetal acceleration \( a \) is expressed as follows:
\( a = \frac{v^2}{r} = r\omega^2 = v\omega \)
From the equation of motion \( F = ma \), the force acting on the object directed toward the center is called the centripetal force. Letting mass be \( m \), the magnitude of the centripetal force \( F \) is:
\( F = m\frac{v^2}{r} = mr\omega^2 \)
Common Mistake: "Centripetal force" is not some brand-new, special type of force that suddenly appears in nature! It is simply the role played by the resultant of the "actual forces acting on the object"—such as tension in a string, normal force from a surface, friction, or gravity—whenever that net force points toward the center of the circle. When solving problems, the golden rule is to always begin by drawing the actual forces acting on the object as usual.
(3) Centrifugal Force (An Fictitious Force Viewed from the Rotating Frame)
From the perspective of an observer rotating along with the object undergoing circular motion, the object appears stationary. In this reference frame, an apparent force directed outward from the center that feels balanced with the centripetal force is called the centrifugal force.
The magnitude of the centrifugal force is equal to the centripetal force, and its direction points radially outward from the center.
\( F_{\text{centrifugal}} = mr\omega^2 = m\frac{v^2}{r} \)
Fun Fact: The spin cycle of a washing machine works by spinning at high speed, exceeding the maximum centripetal force available to hold water droplets to the fabric, which slings the droplets off in the tangential direction. From the rotating frame of reference, this can be explained as "centrifugal force throwing the water outward."
Section 1 Summary
The basic formulas for circular motion are \( v = r\omega \), \( a = \frac{v^2}{r} = r\omega^2 \), and \( T = \frac{2\pi}{\omega} \). When analyzing from an inertial (stationary) frame, set up the "equation of motion \( ma = F \) (toward the center)"; when analyzing from a rotating frame, set up a "balance of forces including centrifugal force."
2. Simple Harmonic Motion (Moving Back and Forth)
Regular, periodic oscillation back and forth along a straight line about a central point is called simple harmonic motion (SHM). At first glance, it may seem complex, but if you visualize it as the shadow of uniform circular motion, all the formulas connect naturally.
(1) Displacement, Velocity, and Acceleration in SHM
Let's consider shining light from the side onto uniform circular motion of radius \( A \) to project its shadow onto a screen.
Let the angular velocity of the circular motion be \( \omega \) (in SHM, this is called the angular frequency), and suppose the object passes through the origin (center) in the positive direction at time \( t = 0 \). The position of the shadow (displacement \( x \)) at time \( t \) is given by:
・Displacement: \( x = A\sin\omega t \)
(where \( A \) is half the total range of oscillation, known as the amplitude)
・Velocity: \( v = A\omega\cos\omega t \)
・Acceleration: \( a = -A\omega^2\sin\omega t = -\omega^2 x \)
This relation \( a = -\omega^2 x \) is the most essential defining feature of simple harmonic motion!
When displacement \( x \) is positive (to the right of the center), acceleration \( a \) is negative (to the left). There is always an acceleration directed toward the center of oscillation.
The period \( T \) (the time for one complete oscillation) and the frequency \( f \) (the number of oscillations per second) are expressed just as in circular motion by the following equations:
\( T = \frac{2\pi}{\omega} \)
\( f = \frac{1}{T} = \frac{\omega}{2\pi} \)
(2) Key Values at Specific Points (Where Quantities Are Maximized or Zero)
A qualitative understanding of "where speed is at a maximum and where acceleration is at a maximum" is tested very often on the Common Test.
・Center of Oscillation (\( x = 0 \))
Displacement: \( x = 0 \)
Velocity: Maximum \( v_{\max} = A\omega \)
Acceleration: \( a = 0 \)
Restoring Force: \( F = 0 \)
・Turning Points / Extreme Ends (\( x = \pm A \))
Displacement: Maximum \( x = \pm A \)
Velocity: Momentarily at rest, so \( v = 0 \)
Acceleration: Maximum \( a_{\max} = A\omega^2 \) (directed toward the center)
Restoring Force: Maximum (directed toward the center)
Think of playing on a playground swing! You move the fastest when passing through the lowest point (the center), and you stop momentarily at the highest points on either end (the turning points). Your physical intuition matches everyday experience perfectly!
(3) Restoring Force
Substituting \( a = -\omega^2 x \) into the equation of motion \( F = ma \) gives:
\( F = m(-\omega^2 x) = -(m\omega^2)x \)
If we define \( K = m\omega^2 \) (a constant), this can be written as:
\( F = -Kx \)
This equation means that there is "a force proportional to the deviation from the center (displacement \( x \)) that always acts to pull the object back toward the center." This force is called the restoring force.
Conversely, whenever an object experiences a net force of the form \( F = -Kx \), that object will always undergo simple harmonic motion! In that case, the angular frequency \( \omega \) and period \( T \) are:
\( \omega = \sqrt{\frac{K}{m}} \)
\( T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{K}} \)
3. Representative Examples of Simple Harmonic Motion
The two classic examples you must master for high school physics are the "mass-spring oscillator" and the "simple pendulum."
(1) Mass-Spring Oscillator
Consider an object of mass \( m \) placed on a smooth horizontal surface, attached to a spring with spring constant \( k \), and set into oscillation.
Taking the displacement from the natural length to be \( x \), the force exerted by the spring (elastic force) is \( F = -kx \).
Since the proportionality constant of the restoring force is simply \( K = k \), substituting this into the formula directly gives the period:
\( T = 2\pi\sqrt{\frac{m}{k}} \)
For a vertically suspended mass-spring oscillator:
When a spring is suspended from the ceiling and a mass is attached, gravity stretches the spring slightly until it hangs at rest. This resting position is the equilibrium position, which becomes the center of oscillation.
Taking the displacement from the equilibrium position as \( x \), the resultant of gravity and the elastic force is still \( F = -kx \), so the object undergoes SHM centered at the equilibrium position. The formula for the period is completely identical to the horizontal case: \( T = 2\pi\sqrt{\frac{m}{k}} \)!
Key Point: For a vertical mass-spring system, the effect of gravity only shifts the center of oscillation to the equilibrium position; the period and the periodic nature of the oscillation do not depend on the acceleration due to gravity \( g \). This is a crucial point that comes up constantly on the Common Test!
(2) Simple Pendulum
Consider a bob of mass \( m \) attached to the end of a string of length \( l \), oscillating through small angles.
Letting the angle from the vertical be \( \theta \) (in radians), the component of gravity acting perpendicular to the string (along the tangential direction) is \( -mg\sin\theta \).
When the angle of swing \( \theta \) is sufficiently small, the mathematical approximation \( \sin\theta \fallingdotseq \theta \) holds. Furthermore, if the arc displacement from the center is \( x \), then \( x = l\theta \), which means \( \theta = \frac{x}{l} \).
Therefore, the restoring force can be approximated as:
\( F = -mg\sin\theta \fallingdotseq -mg\theta = -mg\left(\frac{x}{l}\right) = -\left(\frac{mg}{l}\right)x \)
Since the restoring force constant is \( K = \frac{mg}{l} \), the period \( T \) is derived as:
\( T = 2\pi\sqrt{\frac{m}{K}} = 2\pi\sqrt{\frac{m}{\frac{mg}{l}}} = 2\pi\sqrt{\frac{l}{g}} \)
Common Mistake: The period of a simple pendulum, \( T = 2\pi\sqrt{\frac{l}{g}} \), contains neither the mass \( m \) of the bob nor the amplitude \( A \)!
In other words, whether you make the bob heavier or lighter, or whether you swing it a bit wider or narrower (as long as it remains small), the period does not change. This property is called the "isochronism of the pendulum." It is a famous principle famously discovered by Galileo Galilei while observing a swinging chandelier in a cathedral.
Tip for Remembering the Formulas
If you ever get confused about what goes inside the square root for the periods of a spring oscillator and a simple pendulum, check the physical units (dimensions) or physical intuition:
・Spring oscillator: \( \sqrt{\frac{m}{k}} \) (a heavier mass \( m \) oscillates more slowly, so \( m \) is in the numerator and \( k \) is in the denominator)
・Simple pendulum: \( \sqrt{\frac{l}{g}} \) (a longer string \( l \) oscillates more slowly, so \( l \) is in the numerator and \( g \) is in the denominator)
4. Final Review Checklist for the Common Test!
Finally, let's review the high-yield check points that will give you an edge on the Common Test:
・The direction of acceleration in uniform circular motion is always toward the center, and its magnitude is \( a = \frac{v^2}{r} = r\omega^2 \). Tangential acceleration is zero.
・Period of circular motion: \( T = \frac{2\pi}{\omega} = \frac{2\pi r}{v} \).
・Maximum values of displacement, velocity, and acceleration in SHM: Maximum displacement is \( A \), maximum velocity is \( A\omega \) (at the center), and maximum acceleration is \( A\omega^2 \) (at the turning points).
・Period of SHM: For a spring oscillator, \( T = 2\pi\sqrt{\frac{m}{k}} \); for a simple pendulum, \( T = 2\pi\sqrt{\frac{l}{g}} \). Make sure you know exactly what each depends on and what it does not depend on!
Circular motion and simple harmonic motion are deeply interconnected topics. Instead of relying purely on rote memorization, approach your problems by focusing on the role of forces: "it rotates because there is a force directed toward the center," and "it oscillates because there is a force restoring it toward the center"!