Differential equation satisfied by the angle made by the pendulum.

Question

A pendulum of mass $M$ and length $l$ is pulled to an angle $\alpha$ from the vertical and released from rest. Write down the differential equation satisfied by the angle $\theta(t)$ made by the pendulum with the vertical at time $t$, using the principle of conservation of energy. (If $s$ is the arc length measured from the vertical position, then the velocity $v$ is given by $v = \frac{ds}{dt}$.

how to think?

My attempt is The gravitation force on the object is $mg(l\cos \theta - l\cos \alpha)$ since pendulum is pulled to an angle $\alpha$ from the vertical and then released. Also $$F = m a = m \frac{dv}{dt} = m \frac{d^2 s}{dt^2}$$ Now since $s = l \theta \implies ds = l d \theta$, we get $$F = ml^2 \frac{d^2 \theta}{dt^2} = mgl(\cos \theta - \cos \alpha)$$ But the answer is $\frac{1}{2}l^2\big(\frac{d\theta}{dt}\big)^2 = gl(\cos \theta - \cos \alpha)$. Where am i wrong? Please help.

Answer 1

Recall that we need to equate the

• Kinetic energy: $\frac12mv^2=\frac{1}{2}ml^2\big(\frac{d\theta}{dt}\big)^2 $

to the

• Potential energy: $mg\Delta h=mgl(\cos \theta - \cos \alpha)$