Suppose I have a mathematical pendulum swinging in a plane, and the angle between the vertical direction and the pendulum is $\phi(t) = \phi_0 \cos\omega_0t$. The length of the pendulum is $l$, its mass is $m$. I have to derive the velocity of the pendulum as the function of $\phi$ for solving another problem.

I have derived \begin{equation}v = l \dot{\phi} = l \omega_0 \phi.\end{equation} However the solution suggests to use energy conservation, namely \begin{equation}mgl(1-\cos\phi_0) = mgl(1-\cos\phi) + \frac{1}{2} m v^2(\phi),\end{equation} which gives a completely different function $v(\phi)$.

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    $\begingroup$ The solution $\phi(t)$ you are proposing is an approximate one and holds for when $\phi$ remains small throughout the motion. Using conservation of energy, on the other hand, gives you a correct result. $\endgroup$
    – Phoenix87
    Dec 28, 2014 at 21:24
  • 1
    $\begingroup$ By the way, you have a mistake, $v = - l \phi _0 \omega _0 sin(\omega _0 t)$ s.t. $v = - l \omega _0 \sqrt(\phi _0^2 -\phi ^2)$ . $\endgroup$
    – Sofia
    Dec 29, 2014 at 1:07
  • $\begingroup$ @Phoenix87 Thanks, this was the key to understand the problem. $\endgroup$ Dec 29, 2014 at 11:30

3 Answers 3


This answer is based on $\phi = \phi_0 \cos(\omega_0 t)$ being true.


It is important to note that $l\dot\phi \neq l\omega_0 \phi$. You find this to be the case whenever you find $\dot\phi$ by differentiation.

$$\dot\phi = \frac{d}{dt}(\phi)=\frac{d}{dt}(\phi_0\cos(\omega_0t))$$ $$= -\phi_0 \omega_0 \sin(\omega_0 t)$$

$$ \therefore v = -l\phi_0 \omega_0 \sin(\omega_0 t)$$

The $\sin(\omega_0 t)$ is slightly problematic when we want to express $v$ as a function of $\phi$ because we still have a function of $t$. Let's find a work around:

$$v^2 = l^2 \phi_0^2 \omega_0^2 \sin^2(\omega_0 t)$$

$$= l^2 \phi_0^2 \omega_0^2 (1-\cos^2(\omega_0 t))$$

$$= l^2 \omega_0^2 (\phi_0^2 - \phi_0^2 \cos^2(\omega_0 t))$$

$$= l^2 \omega_0^2(\phi_0^2 - \phi^2)$$

Therefore, the magnitude of the velocity is given by:

$$|v| = l \omega_0 \sqrt{\phi_0^2 - \phi^2}$$


However, as Phoenix87 points out, this is only an approximation for small $\phi$. This means that the equation for the conservation of energy will only be equal if you also approximate the conservation of energy equation:

$$1 - cos(x) \approx \frac{1}{2} x^2 $$

$$\therefore mgl(\frac{1}{2} \phi_0^2) = mgl(\frac{1}{2} \phi^2) + \frac{1}{2} mv^2$$ $$v^2 = gl(\phi_0^2 - \phi^2)$$ $$|v| = \sqrt{gl}\sqrt{\phi_0^2 - \phi^2}$$ Note that from first principles (see below), that, by approximation, $\omega_0^2 = g/l$, so:

$$|v| = l\omega_0 \sqrt{\phi_0^2 - \phi^2}$$ So we get the same expressions for the magnitude of $v$.


To see where the approximation of $\phi = \phi_0 \cos(\omega_0 t)$ comes, we could look at the derivation of the differential equation for $\phi$:

The torque, $\tau$, about the point of attachment of the pendulum to the ceiling is given by:

$$\tau = -mgl \sin(\phi)$$

The rotational variation of Newton's 2nd Law is:

$$\tau = J \ddot \phi $$

where $J$ is the moment of inertia about the point of attachment of the pendulum (analogous to the mass in $F = ma$).

$$J = ml^2$$

$$\therefore -mgl\sin(\phi) = ml^2\ddot \phi$$

So, we get:

$$l\ddot \phi + g \sin(\phi) = 0$$

This is not a nice equation to solve at all, so... we linearise! We assume $\phi$ to be small, so $\sin(\phi) \approx \phi$.


$$l\ddot \phi + g \phi = 0$$

And, if you solve this the way you solve other linear second order differential equations, you get:

$$\phi = A\sin\left(\sqrt{\frac{g}{l}}t\right) + B\cos\left(\sqrt{\frac{g}{l}}t\right)$$

where $A$ and $B$ are arbitrary constants that we will determine by two boundary conditions, where the pendulum is initially held at rest at angle $\phi_0$:

$$t = 0, \phi = \phi_0$$ and $$t = 0, \dot \phi = 0$$

When we substitute the first boundary condition, we get:

$$\phi_0 = B$$

To make use of the second boundary condition, we differentiate the expression for $\phi$ wrt to $t$ to get:

$$\dot\phi = A\sqrt{\frac{g}{l}}\cos\left(\sqrt{\frac{g}{l}}t\right) - B\sqrt{\frac{g}{l}}\sin\left(\sqrt{\frac{g}{l}}t\right)$$

So, substituting the second boundary condition gets us:

$$0 = A\sqrt{\frac{g}{l}}$$

$$\therefore A = 0$$

We have now determine values for the two arbitrary constants, so we finally get the approximate expression for $\phi$ that you started off with:

$$\phi = \phi_0 \cos\left(\sqrt{\frac{g}{l}} t\right)$$

For the sake of convenience, we let $\sqrt{\frac{g}{l}} = \omega_0$ (hence the use of $\omega_0^2 = g/l$ above).

$$\phi = \phi_0 \cos(\omega_0 t)$$

  • $\begingroup$ Please update your answer with the comment of Phoenix87 so that I will be able to accept it. $\endgroup$ Dec 29, 2014 at 11:32
  • $\begingroup$ Done, with the inclusion of some extra points. $\endgroup$
    – Involute
    Dec 29, 2014 at 15:06

If the motion of the pendulum is given to be that of simple harmonic oscillation, then the derivative of the position is the velocity:

$$\phi=\phi_0\cos(\omega t)\\ v = \ell\dot\phi = -\ell\omega\phi_0\sin(\omega t)$$

Which is not the expression you got (you were expressing $v$ in terms of $\phi$ but somehow the $\cos$ became a $\sin$ which is clearly wrong... You can use the trig identity $\sin^2 + \cos^2=1$ to obtain

$$v^2 = (\ell\omega)^2\left(\phi_0^2 - \phi^2\right)\tag1$$

but that is a bit awkward and you end up losing track of the sign of the velocity. But it does accurately reflect conservation of energy... read on.

A simple diagram relating the height and angle shows:

enter image description here

$$h = \ell(1-\cos\phi)$$

So conservation of energy gives

$$\frac12mv^2 + mg\ell(1-cos\phi) = mg\ell(1-cos\phi_0)\tag2$$

Rearranging, we get

$$v^2 = 2g\ell(cos\phi-\cos\phi_0)$$

For small angles $\phi$, we know that $\cos\phi \approx 1-\frac12\phi^2 + O(\phi^4)$ so we rewrite for small angles

$$v^2 = 2g\ell(\frac12\phi_0^2 - \frac12\phi^2)\\ =g\ell(\phi_0^2-\phi^2)\tag3$$

Finally, we know that for small angle, $\omega^2 = \frac{g}{\ell}$ so we can rewrite (1) as

$$v^2 = \frac{g}{\ell}\ell^2(\phi_0^2-\phi^2)\\ =g\ell(\phi_0^2-\phi^2)$$

Which is the same as expression (3). Thus, the derivative of the angular displacement (done correctly) gives the same result as conservation of energy - for small angles, which is assumed by the expression of the motion as a $\cos$ function.

  • $\begingroup$ Yep, thank's for pointing this out, it is a good supplement of Eternal Code's answer. $\endgroup$ Dec 31, 2014 at 9:58

The solution $\phi(t)$ you are proposing is an approximate one and holds for when $\phi$ remains small throughout the motion. Using conservation of energy, on the other hand, gives you a correct result.


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