Error propagation for Atwood’s machine, by simulation

A few weeks ago I mentioned that the propagation of errors is a bit tricky. Say we want to predict the acceleration in an Atwood machine. The machine consists of a string extended over a pulley with masses at either end, of masses M and m, with M > m. The acceleration is given by

a = g{M-m \over M+m}

where g is the acceleration due to gravity, which we assume is known exactly. Let’s set g = 1, so we’ll have

a = {M-m \over M+m}.

We previously found by analytic methods that if M = 100 \pm 1 and m = 50 \pm 1, then a = (1/3) \pm 0.01. But it’s instructive to do a simulation.

Specifically, fix some large n. For i = 1, 2, \ldots, n, let M_i be normally distributed with mean 100 and standard deviation 1; let m_i be normally distributed with mean 50 and standard deviation 1; and let a_i = (M_i-m_i)/(M_i+m_i). Then the mean and standard deviation of the a_i are estimates of the expected acceleration and its error.

This is very easy in R:

n = 10^4;
M = rnorm(n, 100, 1);
m = rnorm(n, 50, 1);
a = (M-m)/(M+m);

When I ran this code I got mean(a) = 0.3333875, sd(a) = 0.00993982. Furthermore, the computed values of a are roughly normally distributed, as shown by this histogram and Q-Q plot. (The line on the Q-Q plot passes through the point (0, mean(a)) and has slope sd(a).)

This works even if the errors are not normally distributed. For example, we can draw the simulated data from a uniform distribution with the given mean and standard deviation:

Mu = runif(n, 100-sqrt(3), 100+sqrt(3))
mu = runif(n, 50-sqrt(3), 50+sqrt(3))
au = (Mu-mu)/(Mu+mu)

I got mean(au) = 0.3332557 and sd(au) = 0.009936519. The distribution of the simulated results is a bit unusual-looking:

There’s also a way to compute an approximation to the error of the result using calculus, but simulation is cheap.


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