An object in simple harmonic motion has an amplitude of 0.202 m and an oscillation period of 0.620 s. Determine the maximum speed of the motion.

Answer 1

T = 2pi*SQRT(L/g) solve for length L of the pendulum

Then, solve for the max angle, knowing L and amplitude

sin theta=Amplitude/L

then compute theta.

Finally, vmax= √{2gL[1-cos(Theta)]}

Answer 2

v(max) =Aω= 2πA/T