$$ T = 2\pi \sqrt{\frac{l}{g}}$$
$$U_{c}(T) =?$$
That's the only thing my teacher left me with. I don't understand how am I supposed to solve this without any measurements.
[Physics] How to calculate uncertainty for a simple pendulum formula?
error analysishomework-and-exercises
Related Solutions
It really depends on if you are allowed to analytically calculate the integral or if you have to do it numerically.
Assuming you are allowed to analytically calculate the integral, you can simplify it as the difference in kinetic energy at the start and end. You have to calculate the velocity only at the beginning and end.
$$E=M\int_{x_\text {start}}^{x_\text {end}}\frac{\mathrm d^2x}{\mathrm dt^2}\mathrm dx=M\int_{v_\text{start}}^{v_\text{end}}v\mathrm dv=\frac{1}{2}M\left(v_\text{end}^2-v_\text{start}^2\right)=\frac{1}{2}M\left(\left(\frac{x_\text{end}-x_{\text{end}-1}}{\Delta t}\right)^2-\left(\frac{x_{\text{start}+1}-x_\text{start}}{\Delta t}\right)^2\right)$$
You can then use the formula to find $\sigma_E^2$ easily.
If, on the other hand, you have to integrate, then you are really doing a sum. You have to calculate every acceleration each time.
$$\int_{x_\text {start}}^{x_\text {end}}\frac{\mathrm d^2x}{\mathrm dt^2}\mathrm dx=\sum_{x_i}\frac{\mathrm d^2x}{\mathrm dt^2}\Delta x=\sum_{x_i}\left(\frac{\mathrm d}{\mathrm dt}\frac{x_{i+1}-x_i}{t_{i+1}-t_i}\right)\left(x_{i+1}-x_i\right)=\sum_{x_i}\frac{\left(\frac{x_{i+1}-x_i}{t_{i+1}-t_i}\right)-\left(\frac{x_i-x_{i-1}}{t_i-t_{i-1}}\right)}{t_{i+1}-t_i}\left(x_{i+1}-x_i\right)\approx\sum_{x_i=x_{\text{start}+1}}^{x_{\text{end}-1}}\left(\left(\frac{x_{i+1}-x_i}{t_{i+1}-t_i}\right)-\left(\frac{x_i-x_{i-1}}{t_i-t_{i-1}}\right)\right)\frac{x_{i+1}-x_{i-1}}{t_{i+1}-t_{i-1}}$$
Propagating error here is going to be much tougher, but only because you're going to have to do a lot of algebra. The first step might be too rewrite the expression so that the sum only involves $x_i$ and not $x_{i+1}$ or $x_{i-1}$.
Here's what you do:
Start with the area formula (no explanation needed): $$ A = lw $$
Add the derivatives in quadrature:
$$ (dA)^2 = (l\cdot dw)^2 + (w \cdot dl)^2 $$
Use the area formula's inverses:
$$ (dA)^2 = \left(\frac A w dw\right)^2 + \left(\frac A l \cdot dl\right)^2 $$
Divide by $A^2$ and take the square root:
$$ \frac{dA}A = \sqrt{\left(\frac{dw}w\right)^2 + \left(\frac{dl}l\right)^2 }$$
So that checks out. This is now in a "relative error form"--that is, e.g., a 10% error in length gives a 10% error in area--makes sense.
So if I do it in meter-Hertz:
$$ v = xf $$
(distance $x$ times frequency $f$). I can safely say:
$$ dv = \sqrt{\left(\frac{dx}x\right)^2 + \left(\frac{df}f\right)^2 }$$
Then I just need to convert $df/f$ to a seconds based result--since it's a relative error, I suspect it should work--a 10% error in frequency should have the same (up to sign and to 1st order) effect as a 10% error in time--and we don't care about sign because we're adding in quadrature:
$$ f = \frac 1 T $$
$$ df = \frac{-dT}{T^2} $$
so that
$$ \frac{df}{f} = \frac{\frac{-dT}{T^2}}{\frac 1 T} = -\frac{dT} T$$
Your professor was right.
Best Answer
It you measure the period directly, then you do not need formula - the error is determined by your measurement device + the reaction time, if you need to switch it on/off (like a chronometer).
On the other hand, if you measure the length of the pendulum, then it is necessary to propagate the error using the equation: $$\Delta T = \left|\frac{\partial T(l,g)}{\partial l}\right|\Delta l + \left|\frac{\partial T(l,g)}{\partial g}\right|\Delta g,$$ where $\Delta T, \delta l, \Delta g$ are the errors in period, pendulum length and the free fall acceleration. Assuming that $g$ is known (you probably do not measure it), the equation reduces to $$\Delta T = \left|\frac{d T(l)}{d l}\right|\Delta l = \pi\frac{\Delta l}{\sqrt{gl}}.$$ You can take as $\Delta l$ the smallest possible measurement on the ruler or whatever you use to measure the length.