MATH 308 Lecture 16

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Maple File

Lecture Notes

Quiz over 3.5 and 3.6 next Wednesday

Application: Spring Oscillation

Mass weighing 2 lbs stretches a vertical spring 6 inches (equilibrium).

Mass is pulled down 3 inches and is given an initial velocity of $-\frac{\sqrt{3}}{8}$.

$\begin{array}{rl}m\frac{{\mathrm{d}}^{2}u}{\mathrm{d}{t}^2}& =-ku\\ u^{″}+\frac{1}{4}u& =0\end{array}$

At $t=0$, $u^{\prime }(0)=-\frac{\sqrt{3}}{8}$. At the same time, $u(0)=-3\text{inches}=-\frac{1}{4}\text{feet}$

The solution is $u=-\frac{1}{4}\sqrt{3}\sin \left(\frac{1}{2}t\right)-\frac{1}{4}\cos \left(\frac{1}{2}t\right)$

Notice the coefficients can be written as $-\frac{1}{2}(\frac{\sqrt{3}}{2}\dots \frac{1}{2}\dots )$ These are $\sin \left(\frac{\pi }{3}\right)$ and $\cos \left(\frac{\pi }{3}\right)$, respectively. So our solution takes the form

$-\frac{1}{2}(\sin \left(\frac{\pi }{3}\right)\sin \left(\frac{t}{2}\right)+\cos \left(\frac{\pi }{2}\right)\cos \left(\frac{t}{2}\right))=-\frac{1}{2}\cos (\frac{t}{2}-\frac{\pi }{3})$

Since cosine is an even function, we can drop the negative sign: $\frac{1}{2}\cos (\frac{t}{2}-\frac{\pi }{3})$

Some properties about this equation:

• Amplitude = $\frac{1}{2}$
• Period = $4\pi$
• Frequency = $\frac{1}{2\pi }$

In general, for any solution of the form $u=c_1\sin \left(\frac{t}{2}\right)+c_2\cos \left(\frac{t}{2}\right)$

• Amplitude = $\sqrt{c_1^2+c_2^2}$
• Shift phase = $\tan \varphi =\frac{c_2}{c_1}$

Suppose we have no damping and the mass is acted on by an external force of $2\cos 0.4t$ pounds, the mass is at equilibrium and is released with no initial velocity.

$\frac{{\mathrm{d}}^{2}u}{\mathrm{d}{t}^2}=-\frac{1}{4}u+2\cos 0.4t$

Maple tells us that the solution is $u=-\frac{200}{9}\cos \left(\frac{t}{2}\right)+\frac{200}{9}\cos \left(\frac{2t}{5}\right)$

The particular solution is $\cos 0.4t$ and the solution to the homogeneous equation is $\cos 0.5t$

Thanks to our trig identities, we can put this in the form $2\left(\frac{200}{9}\right)(\sin 0.9t\sin 0.1t)$. The first sine has a period of $2\pi$, while the second has a period of $20\pi$. These cause an interference with each other, creating a very cool-looking plot: (the blue curve is the solution, and the red curves are the $\pm \frac{3200}{9}\cdot 2\sin \left(\frac{t}{20}\right)$ amplitude and frequency envelopes:

300