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 $-{\frac {\sqrt {3}}{8}}$.

${\begin{aligned}m{\frac {\mathrm {d} ^{2}u}{\mathrm {d} t^{2}}}&=-k\,u\\u''+{\frac {1}{4}}u&=0\end{aligned}}$

At $t=0$, $u'(0)=-{\frac {\sqrt {3}}{8}}$. 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 $-{\frac {1}{2}}\left({\frac {\sqrt {3}}{2}}\dots {\frac {1}{2}}\dots \right)$ These are $\sin {\left({\frac {\pi }{3}}\right)}$ and $\cos {\left({\frac {\pi }{3}}\right)}$, respectively. So our solution takes the form

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

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

Some properties about this equation:

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

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 = $\tan {\phi }={\frac {c_{2}}{c_{1}}}$

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 $2\left({\frac {200}{9}}\right)\left(\sin {0.9t}\sin {0.1t}\right)$. 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: