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:




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

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:

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

  • 1
    $\begingroup$ 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.) $\endgroup$ – 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.

  • $\begingroup$ This is the key answer. There is no point to try to find time related quantities from purely energy conservation. $\endgroup$ – 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!

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.