I have been given the following function to model the behaviour of a simple physical pendulum:
$$\theta(t) = \theta_o e^{-\frac{t}{\tau}} \cos\left(2\pi \frac{t}{T} + \alpha\right)$$
Where $\alpha$ is a constant, $T$ is the pendulum's period, and $\tau$ is a time constant of decay. I only understand $\tau$ to be a friction constant of the pendulum, but I am unsure how to measure it experimentally. How would this variable be simply measured in a pendulum with known $T$, mass, length, initial angle etc…
Thanks!
Best Answer
This is similar to damped harmonic motion (where the damping is caused by air resistance and friction). The cosine term in your equation represents the oscillatory motion and the exponential part of the equation ("modulates") determines the decay of the amplitude over time. Assuming you have access to one, a sonic motion detector can be used to create plots of position and velocity as functions of time. If you plot position versus time, this would be a step in the right direction. Using the data obtained from the detector, you should plot an exponential fit of the decay of this oscillating system. Using your equation you can get the ratio of two successive peaks which would give (and I have ignored your original phase $\alpha$ since we can measure from the start of the first oscillation)
$\large \frac{\theta_1}{\theta_2} =\frac{ \theta_o e^{\frac{-t}{\tau}} \cos(2\pi \frac{t}{T})}{ \theta_o e^{\frac{-(t + T)}{\tau}} \cos(2\pi \frac{(t+T)}{T})}$
and because the motion is periodic the cosine terms are essentially equal (successive peaks), the $\theta_0$ cancel and use the rules for dividing exponents, we get
$\large \frac{\theta_1}{\theta_2}= \large e^{\frac{T}{\tau}}$
If we take the natural log of both sides and rearrange we get,
$\large \tau = \frac{T}{\large log \frac{\theta_1}{\theta_2}}$
where T is the period between each oscillation. If you apply this equation to each pair of peaks in a given time interval, and do this as many times as required for accuracy you will get an average of value of $\tau$.