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.
A
1.5ms-1
B
3.0ms-1
C
4.5ms-1
D
6.0ms-1
Verified Correct Answer: A.
Physical Principles & Numerical Solution
V\(_{max}\) = \omegar
\omega = 2\(\pi\)f but f = \(\frac{1}{T}\)
\omega = \(\frac{2\pi}{T}\) = \(\frac{2\pi}{4\pi}\) ( since T = 4\(\pi\) given)
\omega = 0.5rads/secs
V\(_{max}\) = \omegar = 0.5 x 3 = 1.5m/s ( r = 3cm given)
Source: Myschool Past Questions Archive (Record #33768)