## Tasks

- Measure the effect of varying mass on a simple pendulum’s oscillation period.
- Determine the power-law relationship between a pendulum’s period and length using a log-log plot.
- Use a fit to your period and length data to determine the constants in the model equation for period vs. length, and compare to the equation below.

## Resources

- string, rod, and clamp
- mass holder (50 gram) and masses
- stopwatch (e.g. www.onlinestopwatch.net)
- ruler and/or meter stick

## Background

show/hide

It can be shown for any system undergoing simple harmonic motion that the displacement at any time is proportional to the acceleration, and that the constant of proportionality is equal to the square of the oscillation frequency, in radians per second. a=-\omega^2 x

For a pendulum consisting of a mass on a string and perturbed from vertical by an angle \theta, the restoring force is the component of gravity that is not counteracted by the tension in the string, and is thus equal to F=ma=-mg\sin\theta \approx -mg\theta \approx -mg \frac{x}{L} from which we can determine the oscillation frequency to be \omega=\sqrt{g/L}.

Thus, the time it takes for a mass m hanging from a string of length L to make a single complete swing – is equal to T=2\pi\sqrt{\frac{L}{g}}

## Power-Law Fit

show/hideWe often use fits to determine the relationship between two quantities x and y. However, in order to choose the correct type of fit (e.g. linear or quadratic fit), we usually need to know ahead of time what the power-law relationship between x and y is (i.e. is it y=Ax? y=Ax^2?). In short, in the equation y=Ax^bwe are trying to determine b. If we take the logarithm of both sides, we find \log y = b \log x + \log A

Thus, by performing a linear fit to \log y vs. \log x we can determine the power law of the relationship between the two quantities.