A mass on a horizontal spring
Consider a mass on a frictionless horizontal surface. Connect a spring to it and set the system into motion and it will oscillate back and forth. How can we fully describe the motion?
Start with the free-body diagram. The normal force is cancelled by the force of gravity, and the only horizontal force is the spring force. Therefore:
ΣF = ma
-kx = ma
The acceleration is the second derivative of the position with respect to time. This gives:
-kx |
= m |
d2x
| ![](../blackline.gif) |
dt2
|
|
d2x
| ![](../blackline.gif) |
dt2
|
|
= - |
k
| ![](../blackline.gif) |
m
|
|
x |
Let's guess at a solution: x(t) = A cos(ωt)
Here A represents the amplitude of the oscillation, and ω is called the angular frequency. Also, a sine works just as well as a cosine.
Taking two derivatives of x(t) gives:
v(t) |
= |
dx
| ![](../blackline.gif) |
dt
|
|
= -Aω sin(ωt) |
a(t) |
= |
d2x
| ![](../blackline.gif) |
dt2
|
|
= |
dv
| ![](../blackline.gif) |
dt
|
|
= -Aω2 cos(ωt) |
a = -ω2 x
Compare this to the equation we derived for the mass on the spring:
a |
= - |
k
| ![](../blackline.gif) |
m
|
|
x |
This tells us that the angular frequency is given by:
ω2 |
= |
k
| ![](../blackline.gif) |
m
|
|
Motion described by an equation like x(t) = A cos(ωt) is known as simple harmonic motion.
Given that a = -ω2 x, any time you find the acceleration to be proportional to the negative of the displacement you'll have a system experiencing simple harmonic motion, with the angular frequency being the square root of whatever is multiplying -x.