Problem — Log-log bounce

Problem

A ball bounces and the following pairs are measured (release height $h$, time of a single bounce $\tau$): $(0{,}20;\ 0{,}20)$, $(0{,}80;\ 0{,}40)$, $(1{,}80;\ 0{,}60)$, $(3{,}20;\ 0{,}80)$, with $h$ in metres and $\tau$ in seconds. Verify graphically (log-log) that $\tau = a\,h^{n}$ and determine $a$ and $n$.

Solution

Idea. If a power law $\tau = a\,h^{n}$ holds, taking the logarithm of both sides gives $\log\tau = \log a + n\log h$ that is, in the variables $\log h$ and $\log\tau$, a straight line of slope $n$ and intercept $\log a$. Plotting the data on a log-log graph should therefore give points that line up.

Estimating the exponent $n$. It is enough to observe how $\tau$ changes when $h$ is multiplied by a known factor. Going from the first to the second pair, $h$ goes from $0{,}20$ to $0{,}80$, i.e. it quadruples ($\times 4$), while $\tau$ goes from $0{,}20$ to $0{,}40$, i.e. it doubles ($\times 2$). Since $4^{n} = 2$ implies $n = 1/2$, the law is of the form $\tau \propto \sqrt{h}$. Checking with the other pairs confirms this: from $h=0{,}20$ to $h=3{,}20$ the factor is $\times 16$ and $\tau$ goes from $0{,}20$ to $0{,}80$, i.e. $\times 4 = 16^{1/2}$. Consistent.

On the log-log graph the slope is $n = \frac{\log\tau_2 - \log\tau_1}{\log h_2 - \log h_1} = \frac{\log(0{,}80/0{,}20)}{\log(3{,}20/0{,}20)} = \frac{\log 4}{\log 16} = \frac{1}{2}$

Estimating the coefficient $a$. From $\tau = a\sqrt{h}$ we get $a = \tau/\sqrt{h}$. Using the first point $(h=0{,}20,\ \tau=0{,}20)$: $a = \frac{0{,}20}{\sqrt{0{,}20}} = \sqrt{0{,}20} \approx 0{,}45 \ \mathrm{s\cdot m^{-1/2}}$

Sketch of the log-log graph (points lying on a line of slope $1/2$):

$\ev{n = \tfrac{1}{2} \qquad a \approx 0{,}45\ \mathrm{s\cdot m^{-1/2}} \qquad \tau \approx 0{,}45\,\sqrt{h}}$

Links

Topics: Experimental method Skills: Using logarithmic scales · Linearising data · Reading graphs