WAEC - Physics (2000 - No. 12)

The period of oscillation of a particle executing simple harmonic motion is 4\(\pi\) seconds. If the amplitude of oscillation is 3.0m. Calculate the maximum speed of the particle.

1.5ms^-1

3.0ms^-1

4.5ms^-1

6.0ms^-1

Explanation

V\(_{max}\) = ωr

ω = 2\(\pi\)f but f = \(\frac{1}{T}\)

ω = \(\frac{2\pi}{T}\) = \(\frac{2\pi}{4\pi}\) ( since T = 4\(\pi\) given)

ω = 0.5rads/secs

V\(_{max}\) = ωr = 0.5 x 3 = 1.5m/s ( r = 3cm given)

WAEC - Physics (2000 - No. 12)

Explanation

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