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Question: The period of a simple pendulum is given by $T=2\pi\sqrt{\frac{L}{g}}$ where $L$ is the length of the pendulum in feet, $g$ is the constant of acceleration due to gravity, and $T$ is measured in seconds. Suppose that the length of a pendulum was measured with a maximum error of $\frac{1}{4}$%.

What will be the maximum percentage error in measuring its period?

So this is what I have so far. $$Δ{t} ≈ dT=T'dT$$ Error in period $$\frac{ΔT}{t} ≈ \frac{dT}{T}$$$$dL=\frac{1}{4}$$ $$T'=2\pi(\frac{\sqrt{g}\frac{1}{2}(L)^{-\frac{1}{2}}-\sqrt{L}\frac{1}{2}(g)^{-\frac{1}{2}}}{g})$$ Annnnnnd that's all. I have no idea how to approach this problem and I don't even know if what I have so far is right. A detailed step by step explanation would be greatly appreciated, thanks in advance.

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The following computes the first order relative error. In this case one can compute the maximum relative error exactly (since $\sqrt{\cdot}$ has nice properties), but ones misses out on seeing the functional form of the sensitivities.

Think of $T$ as a function of $L$, that is, $T(L) = 2 \pi \sqrt{L \over G}$.

Then a small change in $L$, say $\lambda$, will result in a change in $T$ of $T(L+\lambda) -T(L) \approx T'(L)\lambda$.

Hence the relative error is $\approx {T'(L) \over T(L)} \lambda$.

In this case you are given $\lambda \approx {1 \over 400} L$, so the relative error becomes $\approx {T'(L) \over T(L)} {1 \over 400} L$.

We have $T'(L) = \pi \sqrt{1 \over LG }$, so substituting gives the relative error $\approx \pi \sqrt{1 \over LG } {1 \over 2 \pi} \sqrt{G \over L} {1 \over 400} L = {1 \over 800}$.

A quick check shows that the quantity is 'unit less', as it should be, and as a percentage, this is ${1 \over 8}$% (as would be expected as the power of $L$ in $T(L)$ is ${1 \over 2}$).

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  • $\begingroup$ Thanks a lot, I'm still having trouble fully grasping the concept, but this helps. I'm sure more practice will help. $\endgroup$ – Kenshin Jul 11 '14 at 3:32
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Brute force: $$ T_1 = 2\pi \sqrt{\frac{L}{g}}\sqrt{1\pm 0.0025} = T_0 \left(1\pm \frac{1}{2}.0025 +\frac{\frac{1}{2}(\frac{1}{2}-1)}{2}0.0025^2+\cdot\cdot\cdot\right) $$ So the error is $\frac{1}{2}0.0025$

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  • $\begingroup$ If you are going to compute the error like this, you might as well compute it exactly rather than using a Taylor expansion. Since $\sqrt{\cdot}$ is concave and strictly increasing at $1$, the maximum relative error will be $1-\sqrt{1-0.0025}$. $\endgroup$ – copper.hat Jul 11 '14 at 3:21
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First you have a measure of the real period $T$ with error, call it $T'$. And a measure of the length $L$ with error call it $L'$ $$T'=2\pi\sqrt{\frac{L'}{g}}$$

Assume that your $L'$ have an error of max $\frac{1}{4}\%$. Then $$L'=L(1+\epsilon)$$ Where $\epsilon$ is the error with $\left|\epsilon\right| \leq \frac{1}{4}\%$

Replacing:

$$T'=2\pi\sqrt{\frac{L'}{g}}=2\pi\sqrt{\frac{L(1+\epsilon)}{g}}=2\pi\sqrt{\frac{L}{g}}\sqrt{1+\epsilon}=T\sqrt{1+\epsilon}$$

Therefore:

$$\left|\frac{T}{T'}-1\right|=\left|\sqrt{1+\epsilon}-1\right|\leq\left|\sqrt{1-\frac{1}{400}}-1 \right|= 0.00125078222809105=0.12507\%$$

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  • $\begingroup$ The maximum relative error, if you want to compute it exactly), occurs for $1-\epsilon$, since $\sqrt{\cdot}$ is concave and strictly increasing at $1$. The error in this case is $\approx$ 0.0012507... $\endgroup$ – copper.hat Jul 11 '14 at 3:19
  • $\begingroup$ @copper.hat thanks, fixed $\endgroup$ – rlartiga Jul 11 '14 at 3:22
  • $\begingroup$ @rlartiga thanks a lot for the help as well! $\endgroup$ – Kenshin Jul 11 '14 at 3:37
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Here's how an experimental physicist would compute this error. Assuming that all the errors in a given measurement are linearly independent (that is to say, one error does not affect another error), then for any given function$$F(x_1,x_2,...,x_N)$$ the standard error $\sigma_F$ is given by $$ \sigma_F^2=\sum_{i=1}^N\bigg(\frac{\partial F}{\partial x_i}\bigg)^2\sigma_{x_i}^2 $$ Since you only have one source of error, for $T(L,g)$ you have $$ \sigma_T=\frac{\partial T}{\partial L}\sigma_L=2\pi\bigg(\frac{1}{2}\bigg)\bigg(\frac{L}{g}\bigg)^{-\frac{1}{2}}\bigg(\frac{1}{g}\bigg)(.0025L)=.0025\pi\sqrt{\frac{L}{g}}=.00125T$$ This result informs you that your most accurate results will be obtained when the period is small.

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  • $\begingroup$ The above gives the variance of $F$ in terms of the variances of the (independent) $x_i$s, and is practically a better measure of error. However, the problem was looking for maximum relative error in which case the worst case first order error would be something like ${ \sum | { \partial F(x) \over \partial x_i} x_i| r_i \over F(x)}$, where $r_i$ is the maximum relative error of $x_i$. In one dimension it amounts to the same thing (with appropriate matching of $r_i$ and $\sigma_{x_i}$), in higher dimensions the standard deviation estimate will be lower (appropriately) than the maximum. $\endgroup$ – copper.hat Jul 12 '14 at 18:44
  • $\begingroup$ What copper.hat says is true! The real value of the above formula is not so much its utility as a first-order error estimation, but more as a way to compare the relative weights of a number of different sources of error, an important outcome for physicists and engineers trying to maximize the use of research dollars while simultaneously minimizing error. The magnitude of a standard deviation within any given variable is no indicator of how said deviation is propagated into the total error nor how the behavior of the total error changes across a wide range of possible measurements. $\endgroup$ – atomteori Jul 15 '14 at 4:35

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