Question

A particle moves in simple harmonic motion with a frequency of 3.80 Hz and an amplitude of 5.50 cm. (a) Through what total distance does the particle move during one cycle of its motion? cm (b) What is its maximum speed? cm/s Where does this maximum speed occur? as the particle passes through equilibrium at maximum excursion from equilibrium exactly halfway between equilibrium and maximum excursion none of these (c) Find the maximum acceleration of the particle. m/s^2 Where in the motion does the maximum acceleration occur? as the particle passes through equilibrium at maximum excursion from equilibrium exactly halfway between equilibrium and maximum excursion none of these

Answer 1

(A) during 1 cycle of motion,

distance travelled = 4 A = 4 x 5.50 cm

= 22 cm

(B) max. speed = A w = 2 x pi x 3.80 x 5.50

= 131.3 cm/s

as the particle passes through equilibrium.

(c) a_max = A w^2

= 5.50 x (2 x pi x 3.80)^2

= 3135.4 cm/s^2

= 31.35 m/s^2

at maximum excrusion from equilibrium.