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Question

A particle moves in simple harmonic motion with a frequency of 3.00 Hz and an amplitude of 5.00 cm.  (a) Through what total distance does the particle move during one cycle of its motion? (b) What is its maximum speed? Where does this maximum speed occur? (c) Find the maximum acceleration of the particle. Where in the motion does the maximum acceleration occur?


Solution

(a) The distance traveled in one cycle is four times the amplitude of motion, or $$\boxed{20.0 \ cm}$$.
(b) $$v_{max} = \omega A = 2 \pi f A = 2 \pi (3.00 \ Hz)(5.00 \ cm)= 94.2 \ cm/s$$
(c) $$a_{max} = \omega^2 A = (2 \pi f)^2 A = [2 \pi(3.00 \ Hz) ]^2(0.05 \ m)= \boxed{17.8 \ m/s^2}$$
This occurs at maximum excursion from equilibrium.

Physics

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