# I need help figuring this error percentage homework problem.

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.

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 ${1 \over 2}$).

• Thanks a lot, I'm still having trouble fully grasping the concept, but this helps. I'm sure more practice will help. – Kenshin Jul 11 '14 at 3:32

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 $\frac{1}{2}0.0025$

• 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 $1-\sqrt{1-0.0025}$. – copper.hat Jul 11 '14 at 3:21

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:

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

• 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... – copper.hat Jul 11 '14 at 3:19
• @copper.hat thanks, fixed – rlartiga Jul 11 '14 at 3:22
• @rlartiga thanks a lot for the help as well! – Kenshin Jul 11 '14 at 3:37

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.

• 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. – copper.hat Jul 12 '14 at 18:44
• 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. – atomteori Jul 15 '14 at 4:35