# Formula for period of pendulum using energy conservation

I'm trying the derive the period of a simple pendulum using energy conservation and without calculus.

I'm doing something wrong which I can't figure out.

I see a lot of other derivations online using calculus which I want to avoid for now.

The pendulum has a length $$L$$ and is displaced by $$\theta$$ from the vertical.

Conservation of mechanical energy:

$$E_U=E_T$$

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

$$v=\sqrt{2gh}$$

$$v=\sqrt{2gL(1-\cos \theta)}=\omega L$$

I have a feeling something is wrong with saying: $$v=\omega L$$ in the last line.

$$2gL(1-\cos \theta)=\omega^2 L^2=\frac{4\pi^2}{T^2}L^2$$

$$T^2g(1-\cos \theta)=2\pi^2 L$$

$$T^2=2\pi^2 \frac{L}{g} \frac{1}{(1-\cos \theta)}$$

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

This would be great if: $$\frac{1}{\sqrt{2(1-\cos \theta)}}$$ approaches $$1$$ as $$\theta$$ approaches zero but it doesn't.

Though I notice that it would work if say: $$v=\omega L\sin \theta$$, but why would that be ?

Is it true that:

$$T=2\pi^2 \sqrt{\frac{L}{g}} \frac{\sin \theta}{\sqrt{2(1-\cos \theta)}}$$

• It looks like you're assuming that $\omega$ is constant. It isn't, which is precisely what calculus is for - calculus deals with quantities that change. (Having said that, I'm sure somebody knows a clever way to do this without doing an integral or solving an ODE, but it would likely be simpler to just learn calculus.) – jacob1729 May 19 '20 at 0:54

Setting $$v = L\omega$$ is fine. Setting $$\omega = 2\pi/T$$ is incorrect. It would only be correct if the pendulum were traveling in a full circle and at a constant speed, neither of which is true for an oscillating pendulum.

Also, your formula for energy conservation $$mgh = \frac{1}{2}mv^2$$ is only true if $$h$$ is the maximum height and $$v$$ is the maximum speed, which do not occur at the same time. The correct way to express conservation of energy for all points in the swing is $$mgh + \frac{1}{2}mv^2 = E = \textrm{constant}.$$ Then, you can say that, at the maximum height, velocity is zero, so $$mgh_{max} = E$$ and, at maximum velocity, the height is zero (if height is defined as the distance above the lowest point in the swing $$\frac{1}{2}mv_{max}^2 = E.$$ Thus, $$mgh_{max} = \frac{1}{2}mv_{max}^2.$$

You can see how $$v=\omega L$$ gets you in trouble because $$\omega$$ is a constant, and the velocity of the pendulum is clearly not constant. In fact, the velocity is largest when the angular displacement is $$0$$ (bottom of the pendulum), and the velocity is $$0$$ when then when angular displacement is largest. Thus, if you use $$\theta=\theta_0\cos(\omega t)$$, the velocity will be in $$\sin\omega t$$.

To be precise: $$v\approx L\frac{\Delta \theta}{\Delta t}$$ so that, using \begin{align} \Delta \theta &= \theta_0\left(\cos(\omega (t+\Delta t))-\cos(\omega t)\right)\, ,\\ &=\theta_0\left(\cos(\omega t)\cos(\omega \Delta t)-\sin(\omega t)\sin(\omega \Delta t)-\cos(\omega t)\right)\, ,\\ &\approx\theta_0\left(\cos\omega t-\sin(\omega t)\Delta t)-\cos\omega t \right)\, ,\\ &=-\omega \Delta t\theta_0\sin(\omega t) \end{align} where $$\cos(\omega \Delta t)\approx 1$$ and $$\sin(\omega \Delta t)\approx \omega \Delta t$$ have been used. Thus you have \begin{align} v=-L\omega\sin(\omega t) \end{align} Once you have this you can use conservation of energy as you suggest: \begin{align} \frac{1}{2}mL^2\omega^2\theta_0^2\sin^2(\omega t)+ mgL(1-\cos\theta) =mgL(1-\cos\theta_0) \tag{1} \end{align} where $$\theta_0$$ is the amplitude, and the RHS is evaluated when the pendulum is at displacement $$\theta_0$$, where its velocity at that point is $$0$$.

Rearranging for small angles \begin{align} mgL(\cos\theta-\cos\theta_0)&\approx mgL\frac{1}{2}(\theta^2_0-\theta^2)\, ,\\ &=\frac{mgL}{2}\theta_0^2(1-\cos^2\omega t)=\frac{mgL}{2} \theta_0^2\sin^2\omega t \end{align} so that (1) becomes \begin{align} \frac{1}{2}mL^2\omega^2\theta_0^2\sin^2\omega t&= \frac{1}{2}mgL\theta_0^2\sin^2\omega t\, ,\\ \end{align} and the result follows.

I'm pretty sure you can't. Conservation of energy will give you a relation between $$\theta$$ and $$\omega$$, but there is no way you can extract anything time related from it unless you write $$\omega=d\theta /dt$$.

If $$\theta_0$$ is the maximum angle, then $$\frac 12 mL^2\omega^2-mgL\cos\theta=-mgL\cos\theta_0.$$ Which yields $$\omega=\pm\sqrt{\frac {2g}{L}(\cos\theta-\cos\theta_0)},$$ where the sign depends on the current direction of the oscillation. To proceed further you need to write $$\omega=d\theta/dt$$, so you can get $$dt=\sqrt{\frac{L}{2g}}\frac{\pm d\theta}{\sqrt{\cos\theta-\cos\theta_0}}.$$ Now, it's easier to compute a quarter period from $$\theta=0$$ to $$\theta=\theta_0$$ so you just have to take the plus sign of the square root: $$\int_0^{T/4}dt=\sqrt{\frac{L}{2g}}\int_0^{\theta_0}\frac{d\theta}{\sqrt{\cos\theta-\cos\theta_0}}.$$ Finally, $$T=4\sqrt{\frac{L}{2g}}\int_0^{\theta_0}\frac{d\theta}{\sqrt{\cos\theta-\cos\theta_0}}.$$ This integral can't be written in closed-form which further confirms there is no way to compute the period without calculus.

• This is the key answer. There is no point to try to find time related quantities from purely energy conservation. – nasu May 20 '20 at 2:59

$$v=\sqrt{2gL(1-\cos \theta)}=\omega L$$ I have a feeling something is wrong with saying: $$v=\omega L$$ in the last line.

You're right to suspect that there is something fishy here.

• This is a correct thing to write if you mean $$\omega$$ is the instantaneous angular velocity of the pendulum when it reaches the bottom-most point of its oscillation.
• But, it is a wrong thing to write if you mean $$\omega$$ is the angular frequency of the harmonic motion.

Let's clarify the difference between the two.

Although you're allergic to calculus, I suppose you're fine with writing down equations of motion for harmonic motion using trigonometric functions. So, I will use them. What's a harmonic motion of some quantity, say $$q$$? It's the oscillations in the values of $$q$$ with the time given by an equation like

\begin{align} q=A\sin{\omega_{S} t} \end{align} where $$A$$ is the amplitude of the oscillations and $$w_{S}$$ is the angular frequency of oscillations and the time period of the oscillations would thus be $$T=\frac{2\pi}{\omega_S}$$.

Now, if you take $$q$$ to be displacement, you can describe the harmonic motion of a particle in a line, such as that of a mass attached to a spring. If you take $$q$$ to be an angular displacement, you and describe the harmonic motion of a particle in angular motion, such as that of a (low amplitude) pendulum.

Finally, there is a little fact that we need to steal from calculus which is that the rate of change of $$q$$ when $$q$$ is at its mean value is given by $$A\omega_S$$ in harmonic motion. This means that when $$q$$ is some linear displacement (like for a mass attached to a spring), the speed of the mass when the mass is at the point around which it is oscillating would be $$A\omega_S$$. And, similarly, in our case, the angular speed of the pendulum when the pendulum is at its bottom-most point would be given by $$A\omega_S$$.

So, if the angular speed of the pendulum at its bottom-most point is $$\omega$$ then we have to write $$\omega=A\omega_S$$where $$A$$ is the amplitude of oscillation, which is the angle of maximum displacement, $$\theta$$. Thus, we get $$\omega_S=\frac{\omega}{\theta}$$

Now, what you calculated was simply the angular velocity of the pendulum at the bottom-most point. You have to use that to find the angular frequency of the harmonic motion to get the time-period $$T=\frac{2\pi}{\omega_S}$$.

Using your result for $$\omega$$, one can write

\begin{align} T&=\frac{2\pi}{\omega_S}\\ &=\frac{2\pi\theta}{\omega}\\ &=2\pi\sqrt{\frac{L}{g}}\frac{\theta}{\sqrt{2(1-\cos\theta)}}\\ \end{align}

The final step is to use the identity $$1-\cos\theta=2\sin^2\frac{\theta}{2}$$ and the approximation $$\sin\theta\approx \theta$$ for small $$\theta$$ (we have to take the small $$\theta$$ approximation because the assumption of the harmonic motion of a pendulum is only valid under this assumption). I will leave this final step for you!

• ... or use $1-\cos\theta\approx \theta^2/2$ directly. – ZeroTheHero May 19 '20 at 4:56