A particle moves in simple harmonic motion with a frequency of 2.80Hz and an amp

Question

A particle moves in simple harmonic motion with a frequency of 2.80Hz and an amplitude of 4.40 cm. (a) Through %/hat total distance does the particle move during one cycle of its motion? _____ cm (b) What is it 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

Explanation / Answer

here

a)

Total distance = 4 * 4.4 = 17.6 cm

b)

let w be the symbol for omega, so that w = 2 *pie* f = 2 * pie * 2.8 = 5.6 * pie
y = A * sin(wt) = 4.4* sin(5.6 *pie*t)
Vmax = dy/dt = Awcos(wt)
Vmax = A*w = 24.64 *pie cm/s

which occurs as the particle passes through the midpoint

c)

a = dv/dt = -Aw^2sin(wt) so the max. acceleration is
Aw^2 = 108.416 *pie^2 cm/s^2

which occur at the extremes of the range of the particle.

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