Designing a Measuring Instrument

A measuring instrument typically shows its value using a needle that swings across a scale. In some cases the needle may oscillate about the correct reading, and it may be some time until it settles down enough for an accurate reading to be made. On a typical bathroom scale, for example, this may take several seconds. In cases where speed and accuracy are important, such a delay may be unacceptable, and instruments should be designed to produce an accurate reading as quickly as possible.

We will model the measuring instrument as a mass on a spring. The equilibrium position represents the quantity to be measured. The mass $m$ and spring constant $k$ are given, but the damping constant $\gamma$ is under our control. The system will start at some value $y(0)=y_0 > 0$ with $y'(0)=0$, and at some time we will take the measurement $y(t)$. We want to be sure that $y(t)$ is within $.01 y_0$ of the equilibrium value 0.

In order to ensure this, we should wait until some time $t_1$, long enough that $\vert y(t)\vert \le .01 y_0$ for all $t \ge t_1$. It is not good enough just to have $\vert y(t)\vert < .01 y_0$ at some time $t$ if this is not always true later, because it would be hard to ensure that the measurement is taken at the right time. We want to choose $\gamma$ to minimize $t_1$.

I claim that the best $\gamma$ is for an underdamped system where the first minimum of $y$ is at $y = -.01 y_0$. We'll find the $\gamma$ value that satisfies this, and then justify the claim.

Recall that the general solution for free underdamped vibration is of the form $y(t) = A \exp(-\lambda t) \cos(\mu t - \delta)$. The quasi-period is $T = 2\pi/\mu$. Note that $y(t+T/2) = - \exp(-\lambda T/2) y(t)$. Thus if $y'(0)=0$ (the first maximum is at $t=0$), the first minimum occurs at $t= T/2$. The value of $y$ at this first minimum is $-\exp(-\lambda T/2) y_0$. So we want $\exp(-\lambda T/2) = .01$, i.e. $\ln 100 = \lambda T/2 = \lambda \pi/\mu$. Let $\beta = \ln(100)/\pi = 1.465871198$. We want $\lambda = \beta \mu$. Since $\lambda = \gamma/(2m)$ and $\mu = \sqrt{4km-\gamma^2}/(2m)$ we have

$$\,\vcenter{\openup.7ex\mathsurround=0pt \ialign{\strut\hfil$... ...t{4km \beta^2/(1+\beta^2)} = 1.652170109 \sqrt{km} \cr\crcr}}\,$$

This is to be compared with $\gamma=2\sqrt{km}$ which would correspond to critical damping.

With this value of $\gamma$, we have $\lambda = \gamma/(2m) = .826085045 \sqrt{k/m}$ and $\mu = \lambda/\beta = .5635454575 \sqrt{k/m}$. Then $T/2 = \pi/\mu = 5.574692532 \sqrt{m/k}$. The reading can be taken some time before this, although the precise time is not so easy to determine. However, as a practical matter we might wait until the first minimum to take the reading, since the first minimum is easy to see by eye.

With $y = A \exp(-\lambda t) \cos(\mu t - \delta)$, we have

$$y'(t) = A \exp(-\lambda t)(-\lambda \cos(\mu t - \delta) - \mu \sin(\mu t - \delta))$$

$$0 = y'(0) = A (-\lambda \cos \delta + \mu \sin \delta)$$

$$\tan \delta = \lambda/\mu$$

We want $y(0) > 0$ so $\cos \delta > 0$. Thus $\delta = \arctan(\lambda/\mu) = .9721248988$. Then since $y(0)= A \cos \delta$ we get $A = 1.77447975 y_0$. Now we want to solve $A \exp(-\lambda t_1) \cos(\mu t_1 - \delta) = .01 y_0$. Numerical methods indicate $t_1 = 4.191611504$.

Now let's see why this $\gamma$ makes $t_1$ a minimum. If we decrease $\gamma$ slightly, the first minimum is below $-.01 y_0$, and we would have to wait until some time after $T/2$ before taking the measurement. On the other hand, if we increase $\gamma$ it will take longer to reach $y = .01 y_0$. So our $\gamma$ should indeed be a minimum. (This isn't a complete proof, but it should make the result plausible)

Here are the graphs of $y(t)$ for $4 \le t \le 7$ in four cases, all with $m=1$ and $k=1$: our value $\gamma = 1.652170109$, $\gamma = 1.6$, $\gamma = 1.7$, and the critical damping case $\gamma = 2$.