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Understanding the physics of a pendulum


You've probably seen a pendulum swing, but have you ever thought about why it is that a pendulum swings at all? Further, why is it that a pendulum only swings for some period of time becoming motionless? It turns out, a simple pendulum is a great means through which we can develop intuition about the conservation of energy and the relationship between gravitational potential energy and kinetic energy -- not to mention oscillations.
Consider a pendulum of mass $m$, attached to a point of rotation by a massless string of length $L$. Physicists call this a simple pendulum because we are making a simplification by ignoring the mass of the string.

The pendulum begins in an initial state:






And at some point, for the sake of this analysis, it reaches a state mid-swing, which we'll call the final state:




Do not be confused by the use of the word "final" here. This is not the final state of the pendulum's swinging motion. It's true final state is zero motion. Rather by final state, we mean the state of the pendulum after some time has elapsed from the initial state. We are free to choose any initial and final state we want. However, as you'll see, I've chosen these states for a very specific reason -- it will make our analysis easier, and in fact, give us more meaningful information. I encourage you to try this by defining different initial and final conditions, however.

From the conservation of energy, we know it must be true that:
$$E_i = E_f $$
Where, $E_i$ is the initial total energy of the pendulum and $E_f$ is its final total energy.

We can describe $E_i$  and $E_f$ more specifically, in terms of  kinetic energy $K$ and potential energy $U$. Therefore, we can write:

$$E_i = K_i + U_i $$
$$E_f = K_f + U_f $$

Consider the definitions of $U$ and $K$.

$$U = mgh $$
Where $m$ is mass, $g$ is the gravitational acceleration with value $9.8 \frac{m}{s^2}$, and $h$ is height.
$$K = \frac{1}{2}mv^2$$
Where $v$ is velocity.

Now, here's the trick. When using the conservation of energy to analyze a phenomenon, it can be in our best interest to choose our initial and final conditions such that they make the math simpler. If we look carefully at the initial and final conditions, we'll see that $K_i =0$ because the initial velocity has been chosen such that it is at a point in time when the pendulum has zero velocity. We don't know the exact time when this is true, but we know from watching a real pendulum that it will swing up to some max height, at which point it will begin to swing back down. For some instantaneous time, its velocity must be zero. Meanwhile, we also know that $U_f=0$ because we've defined the final condition at a point in time when the pendulum has zero height. As a result, the conservation of energy says:

$$U_i = K_f$$
$$mgh = \frac{1}{2}mv^2$$

The next problem that must be solved is determining the height $h$ of the pendulum.





From this diagram, we know that $L = h + z$ and that $z = Lcos\theta$. We can write the height $h$ as:

$$h = L(1-cos\theta)$$

Referring back to our conservation of energy statement, it follows that:

$$gL(1-cos\theta) = \frac{1}{2}v^2$$

Consider the velocity $v$. This is a special type of velocity because of the semicircular motion the pendulum traces out as it swings.



Indeed, this velocity is a tangential velocity $v_t$. And we know that $v_t = \omega R$, where $R$ is the radius of the circle and $\omega$ is the angular velocity. In this case, $R = L$.

We can re-write our equation as:

$$gL(1-cos\theta) = \frac{1}{2}\omega^2L^2$$

Solve for $\omega$:

$$\omega = \sqrt{\frac{2g}{L}(1-cos\theta)}$$

If we remember that $\omega = 2\pi f$, then we can develop some intuition for what this equation means. It tells us that the motion of the pendulum has a frequency $f$ associated with it. This knowledge combined with our understanding of the relationship between kinetic energy and potential energy suggests that its motion has an oscillation-like nature. Remember: the total energy of a particle consists of kinetic and potential energy. When one increases, the other decreases, and vice versa.

The pendulum swings back and forth, with each swing its gravitational potential energy converted to kinetic energy. In a perfect vacuum, this swinging motion would continue forever. However, on earth -- and in any room with air -- some of the kinetic energy is lost to the surrounding air, as the pendulum collides with air molecules and transfers some of its energy to those molecules. This explains why the pendulum gradually loses height after each period of swinging.

Imagine our pendulum swinging back and forth. At the beginning of each fall, the pendulum has some amount of potential energy; it is converted entirely to kinetic energy and moves downward. But not all of that kinetic energy is returned to potential energy because it is transferred in those collisions with air molecules. As a result, at the end of its upswing, the pendulum reaches a lower height. Some of the motion of the pendulum has been given to the air molecules, so the pendulum is less energetic overall.  This leaking of energy to the surrounding air continues with each period of swinging. Given enough time, the pendulum runs out of energy and becomes motionless.

Without considering the surrounding air, it looks like the energy has disappeared and the conservation of energy has been violated. But it is not so: the energy has been conserved -- it has been transferred outside of the system of the swinging pendulum.

Perhaps a question has entered your mind: how does the height  that the pendulum is dropped from relate to its speed and kinetic energy?

Using just the first two terms of  the Taylor series expansion, we can approximate $cos\theta$ as:

$$ cos\theta = 1 - \frac{\theta^2}{2} $$

Remember, this is an approximation, as we are removing many of the terms from the Taylor series expansion. However, if we use this value and plug it into our equation, we find that:

$$\omega = \theta \sqrt{\frac{g}{L}}$$

The angle $\theta$ is directly related to the height $h$. The greater its value, the higher the pendulum starts its swinging at. This means the greater height from which we drop the pendulum, the greater its frequency of swinging. Further, because the angular velocity $v_t = \omega R$, the initial kinetic energy of the pendulum is greater since it depends on velocity.

As you can see, the pendulum is a fantastic model for understanding the conservation of energy, the relationship between gravitational potential energy and kinetic energy, and oscillation behavior.

Image courtesy Shareef Jackson

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