- #1
saadhusayn
- 22
- 1
We have a car accelerating at a uniform rate ## a ## and a pendulum of length ## l ## hanging from the ceiling ,inclined at an angle ## \phi ## to the vertical . I need to find ##\omega## for small oscillations. From the Lagrangian and Euler-Lagrange equations, the equation of motion is given by
$$l \ddot{\phi} + a \cos(\phi) + g \sin(\phi) = 0$$
Equivalently,
$$l \ddot{\phi} + \sqrt{a^2 + g^2} \sin(\phi + \arctan(\frac{a}{g})) = 0$$
I know that ##\omega = \sqrt{\frac{\sqrt{a^2 + g^2}}{l}}## .
In order to make the small angle approximation for the sine, we must assume that ##a << g##. So is it the case that ##\omega = \sqrt{\frac{\sqrt{a^2 + g^2}}{l}}## only if ##a << g##, because this assumption is not stated in the question.
$$l \ddot{\phi} + a \cos(\phi) + g \sin(\phi) = 0$$
Equivalently,
$$l \ddot{\phi} + \sqrt{a^2 + g^2} \sin(\phi + \arctan(\frac{a}{g})) = 0$$
I know that ##\omega = \sqrt{\frac{\sqrt{a^2 + g^2}}{l}}## .
In order to make the small angle approximation for the sine, we must assume that ##a << g##. So is it the case that ##\omega = \sqrt{\frac{\sqrt{a^2 + g^2}}{l}}## only if ##a << g##, because this assumption is not stated in the question.