# Oscillation period of an ideal pendulum (help with differential calculus)

I'm a first year physics major student, and this is my first question here.

It's a well known fact that ideal pendulums with the same gravity acceleration and same length have the same period, now, I'm trying to proof it (and I might need some help with this). The goal is to make an experiment at home and compute the mass of the Earth (knowing that the period depends on the lenght and gravity only).

I'm aware that the method is also well-known, but it's ok if I want to 'check' what I have? I mean, I really want to to do this.

Let's go! I first drawed something like this;

An ideal string of lenght 'l' and a particle of mass 'm', the angular speed vector is shown as coming out of the page and I defined a coordinate system with the radial and tangencial directions.

Since the particle moves in circular motion and there are only two forces acting on it, then $$\vec{T}+m\vec{g}=m\vec{a}$$ becomes $$-T\hat{u_r}+mg\sin{\theta}\hat{u_t}+mg\cos{\theta}\hat{u_r}=m(\alpha l\hat{u_t}+(-w^2l)\hat{u_r})$$ Where $\hat{u_t}$ and $\hat{u_r}$ are the two directions of my coordinate system.

This way, by projecting on $\hat{u_t}$ we have that $$mg\sin{\theta}=m\alpha l$$ and by solving $\alpha$: $$\alpha=\frac{g}{l}\sin\theta$$

My question is this one: we know that $$\vec{\alpha}=\dot{\vec{\omega }}$$ So, can I do this? $$\alpha=\frac{d\omega}{dt}$$ Implies that $$d\omega=\alpha dt=\alpha dt \frac{d\theta}{d\theta}$$ And since $$\omega=\frac{d\theta}{dt}$$ Is this true? $$\omega d\omega=\alpha d\theta$$ Because if that so, then I can integrate this, right? $$\int_{\omega_0}^{\omega}\omega d\omega=\int_{\theta_0}^{\theta}\frac{g}{l}\sin{\theta}d\theta$$ $$=\frac{\omega^2}{2}-\frac{\omega_0^2}{2}=\frac{g}{l}(\cos{\theta_0}-\cos{\theta})$$

Second question, these $\omega_0$ and $\theta_0$ values must be related? (that is, that they both are from the very same moment), and, can we pick this moment to be that of the maximum amplitude of the pendulum? (so $\omega_0=0$) That way I could have $\omega$ as a function of $\theta$ and maybe use that to find the period.

Well, I hope this not to be annoying in any form to anyone and I'm sorry if the question is a little long.

• Are you trying to find a simple expression fot T? The way you are going will yield the non-elementary elliptic integral. – Triatticus Jul 17 '18 at 19:15
• @Triatticus Thank you! How would you do that? Do you have any source explaining that process? For what I know T 'should' be simple equation, right? – Jesús Isea Jul 17 '18 at 19:15
• @JesúsIsea The only way you can get a simple expression for the period is if you use the small-angle approximation ($\sin\theta\approx\theta$), and this is what is done in most introductory physics courses. If you don't use that approximation, then you get an integral that has no antiderivative in terms of elementary functions. It must be numerically integrated, and the function that you get from this numerical integration is called the elliptic integral. – probably_someone Jul 17 '18 at 19:20
• @probably_someone And this $\sin{\theta}\approx\theta$ can be used directly on the last expression I arrived? (I mean, when I integrate it to find $\theta$) – Jesús Isea Jul 17 '18 at 19:24
• @JesúsIsea Not quite. For one, the expression should be $\alpha=-\frac{g}{l}\sin\theta$. – probably_someone Jul 17 '18 at 19:31

Effectively what you have found is the analogue of the $v\frac{dv}{dx}$ rule - a chain-rule trick - for a rotating system. More streamlined, this is what you have done:

$$\alpha = \frac{d\omega}{dt} = \frac{d\omega}{d\theta} \cdot \frac{d\theta}{dt} = \omega \frac{d\omega}{d\theta} = -\frac{g}{l}\sin\theta.$$

As you mention, you can then integrate both sides to find that

$$\frac{1}{2}\omega^2 - \frac{1}{2}\omega_0^2 = -\frac{g}{l}\int_{\theta_i}^{\theta_f}\sin\theta d\theta = \frac{g}{l}\left[\cos\theta_f - \cos\theta_i\right].$$

This is a statement of energy conservation for the system. Note that if you were to multiply by a moment of inertia $I = ml^2$ to both sides, you would end up with dimensions of energy (i.e., $\frac{1}{2}I\omega^2$ and $mgl\cdot \cos\theta$) on both sides of the equation. This result tells you the change in rotational speed $\omega-\omega_0$ in passing through some angles $\theta_i$ and $\theta_f$.

Unfortunately what this result does not do is give you the velocity $\omega(t)$ or the path $\theta(t)$, i.e. the solution to your original differential equation

$$\alpha = -\frac{g}{l}\sin\theta.$$

It is important to note that $\alpha$ is an angular acceleration, and thus

$$\alpha = \frac{d^2\theta}{dt^2} = \ddot \theta.$$

Therefore,

$$\ddot \theta + \omega_0^2\sin\theta = 0,$$

where I have defined $\omega_0 = \sqrt{\frac{g}{l}}$. To my knowledge, this equation does not have an analytical solution as it stands. However, in the limit of small oscillations, $\sin\theta \approx \theta$ and you have

$$\ddot \theta + \omega_0^2\theta = 0.$$

This can be solved using the auxiliary equation method, replacing orders of differentiation with algebraic powers:

$$u^2 + \omega_0^2 = 0$$ implies that $$u = \pm i\omega_0.$$ A pair of imaginary roots then indicates the solution $$\theta(t) = c_1\sin\left(\omega_0 t\right) + c_2\cos\left(\omega_0 t\right)$$ or something of that family (sines, cosines, complex exponentials). Here $\omega_0$ is the angular frequency and you can easily show that the oscillation period is

$$T = \frac{2\pi}{\omega_0} = 2\pi\sqrt{\frac{l}{g}},$$

as long as we assume the rod has no mass ($I = ml^2$).