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I have the question

The pendulum of a grandfather clock oscillates once every 2.0 seconds. Calculate its acceleration when it is 50mm from the midpoint and has a frequency of 0.5 Hz.

So I have used the equation:

$$ a = -\omega^2 x = \ddot{x}\\ \omega = 2\pi f $$ Therefore $\omega = \pi$.

$x = 50 \text{mm} = 0.05 \text{m}$

Therefore $$ a = -\pi^2 \cdot 0.05 \text{m} $$

Therefore $$ a = -0.5 \text{ms}^-2. $$

Is this correct ?

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It is not correct because $a=\omega^2r$ is valid for uniform circular motion. This would be correct if the pendulum bob were making complete circles at 0.5 Hz and the radius were 50 mm. The question refers to a much longer pendulum that does not make a complete revolution, a seconds pendulum that is just under a meter long. The 50 mm is the horizontal distance from the bottom of the swing to the point where the acceleration is desired.

There is not enough information to answer the question. We need the amplitude of vibration. If the amplitude is 50 mm, the acceleration at that point is just gravity projected on the slope of the path as the velocity is zero. If the amplitude is much larger, the velocity will be larger and the acceleration will have a radial component of $\omega^2r$ as well as the tangential component of gravity projected on the path.

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    Okay I understand now. So I used the wrong equation. So should the correct answer be 0.5 ms^-2 ? – 2017-01-05
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    $a=\omega^2r$ is valid for radial acceleration if you use the current value of $\omega$, but as I pointed out we don't have enough information to compute that, so there is no answer to the question as asked. Using the known length of the pendulum we can compute the tangential acceleration because we know the slope of the path at that point, but we can't get the radial component. – 2017-01-05