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I have a spring system, whose position equation is $$x(t) = c_1cos(8 \sqrt{2}t) + c_2sin(8 \sqrt{2}t)$$

The textbook says it will have a period of motion of $\frac{2 \pi}{(8 \sqrt{2}t)}$. I understand roughly what a period of motion is (it's the time it takes for the system to make a complete cycle) but how is the figure $\frac{2 \pi}{(8 \sqrt{2}t)}$ determined?

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2 Answers 2

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The cosine and sine functions are periodic with period $2\pi$. So set $2\pi = 8\sqrt2 t$ to get $t = \frac{2\pi}{8\sqrt 2}$. At this time, $x(t) = c_1 cos(2\pi) + c_2 \sin(2\pi) = x(0)$, so we're back at the beginning.

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The formula for a $\sin$ or $\cos$ function is $x(t) = A \cos( \omega )$, $x(t) = A \sin( \omega)$

$\omega$ = angular velocity, A - Amplitude (don't care to solve our problem)

$\omega = 2 \frac{\pi}{T}$ , where $T$ is the period

so, the the period for our first function $c_1\cos(8\sqrt{2} * t)$ is $T_1 = 2\pi/\omega \quad \Rightarrow\quad T_1 = 2\pi/(8\sqrt{2}t)$ and we do the same for the sin function $\Rightarrow\quad T_2 = 2\pi/(8\sqrt{2}t)$

Now to find the period for $x(t) = c_1 \cos(8\sqrt{2}t)+ c_2 \sin(8\sqrt{2}t)$ we need to take the Least Common Multiple $(T_1,T_2) = 2\pi/(8\sqrt{2}t)$

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I think that should be $cos(\omega t)$ and $sin (\omega t)$. The argument for a $sin$ and $cos$ function need to be dimensionless always. –  Kitchi Nov 14 '12 at 10:48

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