Calculating acceleration of a pendulum

Question

I have the question "The pendulum of a grandfather clock oscillates once every $2.0$ seconds. Calculate It's acceleration when it is 50mm from the midpoint."

I have used the equation:

$$a = -\omega^2X$$

The final answer I got is $a = -0.5 \frac{m}{s^2}$.

Is this correct ?

Answer 1

A few things you need to know:

1. The period $T$ is related to the angular frequency $\omega$ by $T\cdot \omega = 2\pi$.
2. In the formula $a=-\omega^2 X$, $X$ represents the distance from the midpoint or equilibrium position of the oscillator.

In your problem, you are given $T$, so you can use the first piece of information above to calculate $\omega$. Then you can take that value of $\omega$ and use it in the second piece of information to find the acceleration when $X=50mm$.

Finally, be careful when rounding. If you round prematurely or too aggressively, you will end up with substantial error in your answer. (In this case, your answer of $-0.5 m/sec$ is very close to the correct result, but seems to be an example of overly-aggressive rounding.)

Answer 2

With T as the period.

$T \approx \pi \sqrt {\frac {L}{g}}$

Solve for $L.$

$a = -g + \frac {v^2}{L}\mathbf u$

Where $\mathbf u$ is centripital.

$\mathbf u = (-\frac {50}{L}, \sqrt {1 - (\frac {50}{L})^2})$