Angular velocity of a simple pendulum

[Felipe Lincoln]

Homework Statement

A pendulum with a light rod of length $l$ with a bob of mass $m$ is released from rest at an angle $\theta_0$ to the downward vertical. Find its angular velocity as a function of θ, and the period of small oscillations about the position of stable equilibrium. Write down the solution for θ as a function of time, assuming that $\theta_0$ is small.

Homework Equations

i) $x=\theta l$
ii) $F=-mg\sin\theta$
iii) $V = mgl(1 - \cos\theta)$
iv) $K+V=E$

The Attempt at a Solution

$F$ was obtained considering the equation i), the potential was obtained by doing $F=-\dfrac{dU}{dx}$ and them using the equation iv) we get the first answer, which is to find the angular velocity. The result is $\omega=\pm\sqrt{\dfrac{2g}{l}(\cos\theta - \cos\theta_0)}$.
But what I can't understand is why isn't the angular velocity equal this: $\omega=\sqrt{g/l}$

 

[FactChecker]

A simple sanity check of $w = \sqrt {\frac g l}$: How can the anglular velocity of a pendulum be a constant? It stops at the peak and is at it's maximum velocity at the bottom. You should also apply a similar sanity check to your calculated result.

 

[tnich]

Homework Statement

A pendulum with a light rod of length $l$ with a bob of mass $m$ is released from rest at an angle $\theta_0$ to the downward vertical. Find its angular velocity as a function of θ, and the period of small oscillations about the position of stable equilibrium. Write down the solution for θ as a function of time, assuming that $\theta_0$ is small.

Homework Equations

i) $x=\theta l$
ii) $F=-mg\sin\theta$
iii) $V = mgl(1 - \cos\theta)$
iv) $K+V=E$

The Attempt at a Solution

$F$ was obtained considering the equation i), the potential was obtained by doing $F=-\dfrac{dU}{dx}$ and them using the equation iv) we get the first answer, which is to find the angular velocity. The result is $\omega=\pm\sqrt{\dfrac{2g}{l}(\cos\theta - \cos\theta_0)}$.
But what I can't understand is why isn't the angular velocity equal this: $\omega=\sqrt{g/l}$

I think you are using the variable $\omega$ for two different things. The equation of motion for simple harmonic motion has the form:
$\theta = A\sin(\omega t +\phi_0)$
Notice that in this equation $\omega$ is a constant rate of change of phase angle $\phi$ with time. This is not the same as $\dot\theta$.