KAHIBARO
Table of Contents
Motion of a Mass on a Spring
A spring oscillation is the repeated back and forth motion of an object attached to a spring. The object moves because the spring pulls or pushes it toward an equilibrium position, which is the position where the spring is neither effectively pulling the mass one way nor the other in the idealized description.
When the mass is displaced from equilibrium and released, the spring force causes it to accelerate back toward the center. After passing through equilibrium, the mass keeps moving because of its inertia, then the spring slows it down, stops it for an instant, and pulls it back again. This repeating motion is one of the most important examples of simple harmonic motion.
The Mass-Spring System
Consider a horizontal surface with a spring fixed at one end and a mass attached to the other. If friction is ignored, the mass can move back and forth along a straight line.
The displacement from equilibrium is usually written as $x$. If the mass is pulled to the right, $x$ is positive. If pushed to the left, $x$ is negative.
For an ideal spring, Hooke's law gives the restoring force:
$$F = -kx$$
Here, $k$ is the spring constant, and $x$ is the displacement from equilibrium. The minus sign shows that the force is always opposite to the displacement.
For an ideal spring, the restoring force is
$$F = -kx$$
The force always points toward the equilibrium position.
Why the Motion Is Harmonic
Using Newton's second law, the force on the mass is also
$$F = ma = m\frac{d^2x}{dt^2}$$
Combining this with Hooke's law gives
$$m\frac{d^2x}{dt^2} = -kx$$
or
$$\frac{d^2x}{dt^2} + \frac{k}{m}x = 0$$
This is the differential equation of simple harmonic motion. It tells us that the acceleration is proportional to displacement and opposite in direction.
The standard form of the solution is
$$x(t) = A\cos(\omega t + \phi)$$
A sine form is equally valid:
$$x(t) = A\sin(\omega t + \phi)$$
Here, $A$ is the amplitude, $\omega$ is the angular frequency, and $\phi$ is the phase constant. These quantities describe how the mass moves, but the central result for the spring system is that the angular frequency depends on the spring and the mass.
Angular Frequency, Period, and Frequency
From the equation of motion, the angular frequency is
$$\omega = \sqrt{\frac{k}{m}}$$
Once $\omega$ is known, the period $T$ and frequency $f$ follow:
$$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{k}}$$
$$f = \frac{1}{T} = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$$
These formulas show two important facts. A stiffer spring, larger $k$, produces faster oscillations. A larger mass, larger $m$, produces slower oscillations.
For an ideal mass-spring oscillator,
$$\omega = \sqrt{\frac{k}{m}}, \qquad
T = 2\pi\sqrt{\frac{m}{k}}, \qquad
f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$$
A larger $k$ makes oscillation faster. A larger $m$ makes oscillation slower.
Position, Velocity, and Acceleration
If the position is
$$x(t) = A\cos(\omega t + \phi)$$
then the velocity is the time derivative:
$$v(t) = \frac{dx}{dt} = -A\omega\sin(\omega t + \phi)$$
and the acceleration is
$$a(t) = \frac{d^2x}{dt^2} = -A\omega^2\cos(\omega t + \phi)$$
Since $x(t) = A\cos(\omega t + \phi)$, we can write
$$a = -\omega^2 x$$
This is another way to recognize simple harmonic motion.
The velocity is greatest at equilibrium, where $x=0$. The acceleration is greatest in magnitude at the turning points, where $x = \pm A$.
Important Positions During the Motion
There are three especially important locations in one cycle of motion.
| Position | Displacement $x$ | Velocity | Acceleration | Force |
|---|---|---|---|---|
| Right turning point | $+A$ | $0$ | maximum to the left | maximum to the left |
| Equilibrium | $0$ | maximum magnitude | $0$ | $0$ |
| Left turning point | $-A$ | $0$ | maximum to the right | maximum to the right |
At the turning points, the mass stops only for an instant before reversing direction. At equilibrium, the spring force is zero, but the speed is largest because the mass has gained the most kinetic energy there.
Vertical Spring Oscillations
A mass can also oscillate on a vertical spring. In that case, gravity stretches the spring downward until the mass reaches a new equilibrium position. The oscillation then happens around this shifted equilibrium point.
If displacement is measured from the equilibrium position, the equation of motion has the same form as for the horizontal case:
$$m\frac{d^2x}{dt^2} = -kx$$
So the angular frequency is still
$$\omega = \sqrt{\frac{k}{m}}$$
Gravity changes the equilibrium position, but it does not change the oscillation frequency of an ideal vertical spring.
For a vertical spring, gravity shifts the equilibrium position, but if displacement is measured from equilibrium, the oscillation still satisfies
$$\omega = \sqrt{\frac{k}{m}}$$
How Initial Conditions Determine the Motion
The amplitude and phase depend on how the motion starts. For example, if the mass is pulled to the right by a distance $A$ and released from rest, then at $t=0$,
$$x(0)=A, \qquad v(0)=0$$
A convenient equation is then
$$x(t)=A\cos(\omega t)$$
If instead the mass starts at equilibrium with maximum speed in the positive direction, a sine function is often more convenient.
Different starts produce different phase constants, but the same spring and mass always give the same angular frequency, provided the motion remains ideal.
Physical Interpretation
Spring oscillations are a balance between inertia and restoring force. The spring tries to bring the mass back to equilibrium, while the mass resists changes in its motion. Because of this interplay, the mass does not simply stop at equilibrium. It overshoots and continues into a repeating cycle.
This system is important because many physical situations behave approximately like a mass on a spring when displaced slightly from equilibrium. The mass-spring oscillator is therefore a basic model for many kinds of oscillatory motion.
Summary Relations
| Quantity | Formula |
|---|---|
| Restoring force | $F=-kx$ |
| Equation of motion | $m\dfrac{d^2x}{dt^2} + kx = 0$ |
| Angular frequency | $\omega=\sqrt{\dfrac{k}{m}}$ |
| Period | $T=2\pi\sqrt{\dfrac{m}{k}}$ |
| Frequency | $f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$ |
| Position | $x(t)=A\cos(\omega t+\phi)$ |
| Velocity | $v(t)=-A\omega\sin(\omega t+\phi)$ |
| Acceleration | $a(t)=-\omega^2 x$ |
The key idea of spring oscillations is that an ideal spring produces a restoring force proportional to displacement:
$$F=-kx$$
This leads to simple harmonic motion with
$$T=2\pi\sqrt{\frac{m}{k}}$$
for a mass $m$ attached to a spring of constant $k$.