An educated guess

Let's guess that the solution looks like a sine function. That seems sensible because a sine wave oscillates back and forth and back and forth just like our mass and spring.

\begin{figure}\centerline{\psfig{file=sine.eps,width=4in}}
\end{figure}

The time it takes to complete one complete oscillation is called the period . Here we've stretched the sine wave so that its period is $T$ instead of $2\pi$. Mathematically, we can do that by writing it as $\cos (\omega t)$, where this new parameter $\omega$ is called the angular frequency. From it, you can get the period $T$ as follows. You know that when the argument of the cosine is $2\pi$ you've gone a complete oscillation so that means that

\begin{displaymath}
\omega T ~=~ 2\pi
\end{displaymath} (1.2)

or $\omega ~=~ 2\pi/T$.

The frequency of oscillation means the number of complete cycles that are executed, per second. Mathematically the frequency $f$ is $1/T$, so

\begin{displaymath}
f ~=~ {1\over T} ~=~ {\omega\over 2\pi}
\end{displaymath} (1.3)

Another thing that's important for this system is that the maximum displacement of $x$. Note that unlike a cosine function, it is not $1$. We'll call the maximum displacement the amplitude of oscillation $A$. We can take care of that by multiplying that cosine by $A$. So we have $x(t) ~=~ A \cos(\omega t)$

This describes oscillations pretty well. There's one additional thing. We could have defined $t~=~ 0$ to be at some other time. This will then shift the argument of the cosine by some constant amount. Call it $\delta$. This is often called the phase lag. So we have $\cos(\omega t + \delta)$. It shifts the maximum of the cosine to the left by a fraction $\delta\over 2\pi$ of a period. That corresponds to a time $\delta T ~=~ {\delta\over \omega}$.

\begin{figure}\centerline{\psfig{file=sine2.eps,width=4in}}
\end{figure}

So finally we write


\begin{displaymath}
x(t) ~=~ A \cos(\omega t +\delta)
\end{displaymath} (1.4)

Note that because $\cos(\alpha + \beta) ~=~ \cos\alpha\cos\beta-\sin\alpha\sin\beta$, we could write

\begin{displaymath}
x(t) ~=~ A (\cos(\omega t) \cos\delta- \sin(\omega t) \sin\delta)
\end{displaymath} (1.5)

But $\delta$ and $A$ are arbitrary. So instead of this complicated expression, we could call $C \equiv A \cos\delta$ and $D \equiv A \sin\delta$, so

\begin{displaymath}
x(t) ~=~ C\cos(\omega t)+D\sin(\omega t)
\end{displaymath} (1.6)

This is an equivalent way of describing this kind of oscillation. In other words, adding a sine to a cosine gives a sine wave but shifted in phase, and altered in amplitude.

josh 2010-01-05