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
The correct answer is: A
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)