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WAEC Physics 2010 Paper

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Question : 11

Total: 50

simple harmonic oscillator has a period of 0.02s and amplitude of 0.25m. Calculate the speed in ms^-1 at the centre of the oscillation

125 $\pi$
25.00 $\pi$
100.00 $\pi$
400.00 $\pi$

Solution:

Concept:

At the centre of oscillation, the speed is maximum and given by

v = \omega A

.

Explanation:

Period

T = 0.02\,\text{s}

and amplitude

A = 0.25\,\text{m}

.

Angular frequency is

\omega = \frac{2\pi}{T}

.

Therefore, maximum speed at the centre is

v = \omega A = \frac{2\pi}{T} \times A

.

Substitute the values:

v = \frac{2\pi \times 0.25}{0.02}

.

This gives

v = 25\pi \,\text{ms}^{-1}

.

Answer:

B.

25.00\pi \,\text{ms}^{-1}

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