# Significant Figures in Physics

A string has linear density $10.0 \cdot 10^{-3} \, \mbox{kg/m}$ and is kept under a tension of 100 N. A sinusoidal transverse wave, with a wavelength of 0.30 m, is traveling in the positive direction of an $x$ axis. What is the frequency of the wave?

We are given $\mu = 10.0 \cdot 10^{-3} \, \mbox{kg/m}$, $\tau = 100 \, \mbox{N}$ and $\lambda = 0.30 \, \mbox{m}$. Furthermore, we know that

$$v = f\cdot \lambda = \sqrt{\frac{\tau}{\mu}}.$$

Hence $$\displaystyle f = \frac{\sqrt{\tau / \mu}}{\lambda} = \frac{\sqrt{(100 \, \frac{\mbox{kg}\cdot\mbox{m}}{\mbox{s}^2}) / (10.0 \cdot 10^{-3} \frac{\mbox{kg}}{\mbox{m}})}}{0.30 \, \mbox{m}} = \frac{\sqrt{100 \cdot 10^{3} / 10.0 \, \frac{\mbox{m}^2}{\mbox{s}^2}}} {0.30 \, \mbox{m}} = 300 \, \mbox{Hz}.$$

I know my calculations are right per se, however, I am unsure about the significant figures. The answer is "supposed" to be 333 Hz. But as we are given $N = 100$ – that corresponds to only one significant figure – isn't 333 Hz simply wrong?

You are both right and wrong.

$N=100$ is quote to 3 significant figures (i.e. $1.00 \times 10^{2}$). If this were to be quoted to 1 significant figure it would be $1 \times 10^{2}$.

Similarly, $10.0 \times 10^{-3}$ is given to 3 significant figures.

However $\lambda = 0.30 m$ is only given to two significant figures (i.e. $3.0 \times 10^{-1}$ m).

Finally, you have made a mistake in your calculation. The numerical answer is 333.33 Hz, but as wavelength is only given to 2 significant figures, it might be best to quote this as 330 Hz or $3.3\times 10^2$ Hz.

• First, I am very thankful for your quick reply! Second, on the fact that N = 100 N is quote to three significant figures: Does that not imply that 330 Hz also is quote to 3 significant figures (which it of course is not, considering our calculations)? Commented Jan 11, 2015 at 12:19
• On the mistake in my calculation: do you mean that I should, say on a report, write $\ldots = 333.33 \, \mbox{Hz}$? How should I then present the "other" answer, i.e. that $f = 330 \, \mbox{Hz}$? I hope that you are able to understand my question. Commented Jan 11, 2015 at 12:23
• @Anton - you made a mistake. The number inside the square root sign should be $10^4$ and when you divide by 0.30, you do not get 300. To indicate 330 is only known to 2 significant figures you would write $3.3\times10^2$. Commented Jan 11, 2015 at 12:32
• Thanks for pointing that out! I edited my post for clarity. Thank you again for your help. Commented Jan 11, 2015 at 12:53