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Simple harmonic motion

Simple harmonic motion. Notice that the position of the dot matches that of the sine wave.

Simple harmonic motion (SHM) is the motion of an object which can be modeled by the following function:

$x=A\sin (\omega t+\varphi )$

or

$x=c_1\cos \left(\omega t\right)+c_2\sin \left(\omega t\right)$

where c₁ = A sin φ and c₂ = A cos φ.

In the above functions, A is the amplitude of the motion, ω is the angular velocity, and φ is the phase.

The velocity of an object in SHM is

$v=A\omega \cos (\omega t+\varphi )$

The acceleration is

$a=-A{\omega }^2\sin (\omega t+\varphi )$

Springs and Hooke's Law

An application of this is the motion of a weight hanging on a spring. The motion of a spring can be modeled approximately by Hooke's Law:

F = -kx

where F is the force the spring exerts, x is the position of the end of the spring, and k is a constant characterizing the spring (the stronger the spring, the higher the constant).

Calculus-based derivation

From Newton's laws we know that F = ma where m is the mass of the weight, and a is its acceleration. Substituting this into Hooke's Law, we get

ma = -kx

Dividing through by m:

$a=-\frac{k}{m}x$

The calculus definition of acceleration gives us

$x^{″}=-\frac{k}{m}x$

$x^{″}+\frac{k}{m}x=0$

Thus we have a second-order differential equation. Solving it gives us

$x=c_1\cos \left(\sqrt{\frac{k}{m}}t\right)+c_2\sin \left(\sqrt{\frac{k}{m}}t\right)$ (2)

with an independent variable t for time.

We can change this equation into a simpler form. By lettting c₁ and c₂ be the legs of a right triangle, with angle φ adjacent to c₂, we get

$\sin \varphi =\frac{c_1}{\sqrt{c_1^2+c_2^2}}$

$\cos \varphi =\frac{c_2}{\sqrt{c_1^2+c_2^2}}$

and

$c_1=\sqrt{c_1^2+c_2^2}\sin \varphi$

$c_2=\sqrt{c_1^2+c_2^2}\cos \varphi$

Substituting into (2), we get

$x=\sqrt{c_1^2+c_2^2}\sin \varphi \cos \left(\sqrt{\frac{k}{m}}t\right)+\sqrt{c_1^2+c_2^2}\cos \varphi \sin \left(\sqrt{\frac{k}{m}}t\right)$

Using a trigonometric identity, we get:

$x=\sqrt{c_1^2+c_2^2}[\sin (\varphi +\sqrt{\frac{k}{m}}t)+\sin (\varphi -\sqrt{\frac{k}{m}}t)]+\sqrt{c_1^2+c_2^2}[\sin (\sqrt{\frac{k}{m}}t+\varphi )+\sin (\sqrt{\frac{k}{m}}t-\varphi )]$

$x=\sqrt{c_1^2+c_2^2}\sin (\sqrt{\frac{k}{m}}t+\varphi )$ (3)

Let $A=\sqrt{c_1^2+c_2^2}$ and ${\omega }^2=\frac{k}{m}$. Substituting this into (3) gives

$x=A\sin (\omega t+\varphi )$

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