# Pedulum Problem

1. May 21, 2004

### Hyperreality

A small mass m is attached to a point O by an inextensible string of length a. The mass is held with the string taut, at the same level as O, and released. Determine the angular velocity of the mass when the string makes an angle $$\theta$$with the downwards vertical. From this relationship, determine the period of one complete swing. How does this time compare with the value of the period of a small pendulum of length a?

My solution is
$$mga = \frac{1}{2}m\omega^2a^2 + mga(1-\cos\theta)[\tex] [tex]\omega^2 = \frac{2g\cos\theta}{a}$$

But the answer says

$$\omega^2=\frac{2g\sin\theta}{a}$$

For the second part on the period. Assuming the answer given is right.
$$T^2=\frac{4\pi^2}{\omega^2}[\tex] [tex]T = \pi\sqrt\frac{2a}{g\sin\theta}[\tex] But instead the answer says [tex]T=7.04\sqrt\frac{a}{g}$$

Can anyone please tell me what I've done wrong?

Last edited: May 21, 2004
2. May 21, 2004

### Staff: Mentor

not SHM

Assuming your definition of the angle θ, I agree with your answer for ω. The book's answer is obviously wrong: plug in θ = 0 and you'd get zero where ω should be maximum.

To find the period, it looks like you tried to apply a relationship for simple harmonic motion, interpreting ω as the angular frequency. This is not correct, and furthermore this large amplitude pendulum does not exhibit simple harmonic motion.

Instead, to find the period you need to integrate over an entire period (or just from π/2 to 0 and muliply by 4). (In any case, your answer should not have θ in it: that's your variable of integration.)