The correct answer is C because:

The potential energy of the bob at the release point described is given by $E_{p} = m g h$ , where m is the mass, g is the acceleration due to gravity (9.8 m s⁻²), and h is the height of the bob above its lowest swing point
At the bottom of the swing, all of this potential energy is converted to kinetic energy: $E_{k} = \frac{1}{2} m {v^{2}}_{m a x}$ , where v_max is the maximum velocity of the bob
Equating potential and kinetic energy, $m g h$ = $\frac{1}{2} m {v^{2}}_{m a x}$
$v_{m a x} = \sqrt{2 g h} = \sqrt{2 \times 9.8 \times 0.1}$ = 1.4 m s⁻¹
To obtain the angular frequency, ⍵, use: $v_{m a x} = ω A$ , where A is the amplitude, which will be the maximum displacement of the bob, 0.28 m
$ω = \frac{v_{m a x}}{A} = \frac{1.4}{0.28} = 5$ rad s⁻¹

A is incorrect as the angular frequency is too low

B is incorrect as the angular frequency is too low

D is incorrect as the angular frequency is too high

If you said A then check whether you have converted all distances to metres.

Damping (OCR A Level Physics)

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