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:
$$T=2\pi^2 \sqrt{\frac{L}{g}} \frac{\sin \theta}{\sqrt{2(1-\cos \theta)}}$$
 A: 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,
$$mgh_{max} = \frac{1}{2}mv_{max}^2.$$
A: 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.
A: 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.
A: 
$$v=\sqrt{2gL(1-\cos \theta)}=\omega L$$
  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!
A: It can be shown geometrically that the pendulum angle is θ = sqrt(2h/L) where h is the height of the mass.  Then the energy equation
(1/2)L^2 ( θ' )^2  +  g L ( θ^2 )/2  =  constant. (θ') is first derivative of θ with respect to  t.
We now take the time derivative of the energy expression and it quickly simplifies to
L θ"  =  - g θ .  θ" is the second derivative of θ with t.
Thomas X. Carroll
tcarroll@keuka.edu
(Sorry for not using MathJax)
