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This is actually a physics question, but the final part of the solution is pure mathematics. I have wracked my brains trying to figure out a more elegant solution, which I know to exist.

There are three points that we know of:

$$\begin{align} x(0) & = 4.1 \\ {x\left(\tfrac T2\right)} & = 5.1 \\ x(T) &= 4.5 \end{align} $$

Here $T$ is the period of motion.

The equation for an underdamped harmonic oscillator is:

$$ x(t) = e^{-at}(A\cos\omega t + B\sin\omega t)$$

I have three points, and three unknowns, so I should be able to solve this system. If my calculations have been correct I have gotten the following equations:

$x(0) = A$, because $\sin(0) = 0, e^0 = 1,$ and $\cos(0) = 1$. So $A = 4.1$. One down.

$x(T/2) = e^{-aT/2} \cdot -A$, because $\cos(\pi) = -1, \sin(\pi) = 0$.

$x(T) = e^{-aT} \cdot A$, because $\cos(2\pi) = 1, \sin(2\pi) = 0$.

But now I am totally stumped. I tried to manipulate the equations in such ways as to somehow remove the period, $T$, but I always cancel out a instead, which is what I want to solve for.

For instance:

$$\ln(A/x(T)) = aT \Leftrightarrow T = \ln(A/(x(T)) \cdot \frac1a,$$ but when I substitute this anywhere, I cancel out the $a$.

Can anybody give me some hint as to how to continue onwards, I am sure there is some manipulation that I am just forgetting that would open this equation up to me.

PS.

Sorry for the formatting, no idea how to use $\LaTeX$ and this is urgent.

MJD
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Kayle of the Creeks
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1 Answers1

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Add one extra unknown that corresponds to the equilibrium position we're interested in, so that the equation becomes $$x(t) = Ae^{-\alpha t} \sin (\omega t + \phi) + x_0,$$ where $x_0$ is the equilibrium position. Solve $x_0$ using the 3 values of $x$ you know, the answer is 4,7250 (kg). PS. we're on the same course, ei kestä kiittää

assburger
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