# How do you do analysis of units?

## Homework Statement

Show the unit analysis of this equation that relates frequency to length (in centimeters) of a pendulum

$f = 5.1L^{-0.5}$

## Homework Equations

$f = \frac{1}{2\pi\sqrt{\frac{L}{g}}}$ ?

## The Attempt at a Solution

$f = \frac{1}{2\pi\sqrt{\frac{L}{g}}}$
solving for L:
L = $\frac{g}{2πf^{2}} * \frac{100 cm}{1 m}$
L = $\frac{g}{f^{2}} * \frac{cm}{m}$
L = $\frac{m/s^{2}}{(s^{-1})^{2}} * \frac{cm}{m}$
L = $\frac{m}{s^{2}}* \frac{1}{s^{-2}} * \frac{cm}{m}$
L = $cm$

$f = \frac{1}{2\pi\sqrt{\frac{L}{g}}}$
$f = \frac{1}{\sqrt{\frac{cm}{m/s^2} * \frac{m}{cm}}}$
$f = \frac{1}{\sqrt{s^{2}}}$
$f = s^{-1}$

Is this the right approach?

I think you are on the write track, but maybe making it more complicated than it is. What the question is getting at is how do you find a formula (starting with eqn 2) that relates frequency to length (eqn 1) and assumes L is in cm. Usually length L would be in meters if using MKS units.

Start with eqn 2, but assume L is given in cm, and convert it to meters.

f = 1/(2*pi*sqrt(L*(1 m)/(100 cm)/g)
f = 1/(2*pi*sqrt(L/(100*g))
f = sqrt(100*g)/(2*pi) * 1/sqrt(L)

If g = 9.8 m/s^2 and pi = 3.14, then

f = 5.0 * L^-0.5

which is almost the same as the original result in eqn1. Maybe some roundoff differences in the constants explain the difference between 5.0 here and the original 5.1 given in eqn 1.

Sometimes you might want to do this rearranging of formulas if say, for example, results you get regularly from the lab come in in mixed units. You could make a formula that handles the unit conversion. However, these days there are also some tools out there that can do calculations with mixed units. All the unit conversion is handled automatically, apps like UnityCalc or even Google.

Hope that helps.

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