Period of Simple Harmonic Oscillators

Introduction

As you may already know the two types of simple harmonic oscillators are springs and pendulums. They have many characteristics such as amplitude, frequency, and period (to name a few). This short piece is about the periods of springs and pendulums.

Investigation

For springs, the equation for the period is $T=2\pi \sqrt{\frac{m}{k}}$. This makes intuitive sense because $T$ increases as $m$ increases because objects with more mass have more inertia and thus the spring will “have a harder time” whipping the mass back and forth and the period increases. The fact that $T$ decreases as $k$ increases also makes some natural sense because as $k$ increases, the stiffness and magnitude of force the spring can put out for the same displacement is much greater as well. With a greater amount of force, the speed at which the block travels back and forth (or up and down) increases and the period decreases.

Now what about the equation for the period of a pendulum $T=2\pi \sqrt{\frac{L}{g}}$? If we were to just look at them side by side it is difficult not to notice their many similarities.

$T=2\pi \sqrt{\frac{m}{k}}$ for Springs

$T=2\pi \sqrt{\frac{L}{g}}$ for Pendulums

The second equation seems awkward and unnatural at first glance. However, if we really think about it and infer what we already know from the first equation the second one makes just as much intuitive sense.

For those of you that are wondering why $m$ is not included, think about what would happen if you increased the mass. Sure, you would increase the difficulty of moving the mass at the length of the string, but you also increase the restoring force of the pendulum (as it depends on the gravitational force).

In a more analytical way, the best, easy explanation I could cook up is by recalling some basics of rotation and torque $\tau = I\omega$. If we have a mass $m$ on the end of a string, the rotational inertia would be $I=mL^2$. The torque would be supplied by $mg\sin(\theta)$ which, multiplied by the lever arm, would result in $\tau=L*mg\sin(\theta)$. Thus our original equation $\tau = I\omega$ can be rewritten as $L*mg\sin(\theta)=mL^2\omega$. It then becomes transparent that $m$ can be cancelled on both sides and thus $\omega$ (a measure of the period, of sorts) is independent of mass.

The above explanation for why mass is not included in the equation for period also tells us why $L$ is. In the case of a pendulum the inertia we are interested in is the rotational inertia and not simply the “regular” inertia. If we look at the equation we derived above $L*mg\sin(\theta)=mL^2\omega$ only one $L$ cancels on both sides because $\tau$ increases with $L$ in a linear fashion but $I$ increases with $L$ exponentially. Thus, as increasing $L$ increases rotational inertia and “difficulty” of moving the mass on our pendulum it makes sense that it is on the numerator in $T=2\pi \sqrt{\frac{L}{g}}$.

As for the $g$ at the denominator, we can really think about it as a more general $a$, the acceleration of masses on any planet, not just on Earth. As we increase the acceleration $a$, the force of the restoring force increases because the restoring force is given by $F_{restoring} = ma\sin\theta$. Thus, the period decreases. This is similar to increasing the $k$ in the formula for period in springs.

Recap

The equations for period for springs and pendulums are $T=2\pi \sqrt{\frac{m}{k}}$ and $T=2\pi \sqrt{\frac{L}{g}}$, respectively. They may seem different but are really telling us the same thing. The variables $m$ and $L$ describe the inertia of the body in linear and rotational terms. The denominators $k$ and $g$ model the strength of the restoring forces.