Question

A 0.05kg particle moves in simple harmonic motion with a frequency of 20.0 Hz and an amplitude of 25.0 cm red 5.00 an (a) Through what total distance does the particle move during five cycle of its motion? (b) What is its maximum speed? Where does that occur? (c) Find the maximum acceleration of the particle. Where does that occur? Please write down the full working and indicate your answers for a) b) c) clearly Paragraph B IEE (a)

Answer 1

Given, ma goosky t=2042 A = 25cm Analysis S.H.M, act) - Alos (wt+4) (a) The total distance d' an the particle move during one Cycle is from X=-A to'x=+A and back to X_E-A so, d= 4A. for 5 cycle of its motion, D = SMA) = 30x25 cm) = sopim so, D = Soom (6) The maximum speed of the particle is Vmax = WA Imax = 200A - Vmax = 2x1 * (25cm) x (2012) - Imax = 3141.5962 (m): Vmax = 31.415962 m/s The particle has max. speed when it passes through the equilibrium position. (c) The maximum. acceleration of the the particle is, = W2A = (206)? (2x15 420) 40.25) - (amax = 3947.84176 m/s2 ama amax The particle has maximum acceleration at the turning points, where it has maximum displacement