A particle moves in simple harmonic motion with a frequency of2.20 Hz and an amp

Question

A particle moves in simple harmonic motion with a frequency of2.20 Hz and an amplitude of 5.90 cm.(a) Through what total distance does theparticle move during one cycle of its motion?
1 cm
(b) What is it maximum speed?
2 cm/s
Where does this maximum speed occur? 3 as the particle passes throughequilibrium
at maximum excursion fromequilibrium    
exactly halfway between equilibrium andmaximum excursion
none of these

(c) Find the maximum acceleration of the particle.
4 m/s2
Where in the motion does the maximum acceleration occur? 5 as the particle passes throughequilibrium
at maximum excursion fromequilibrium    
exactly halfway between equilibrium andmaximum excursion
none of these 3 as the particle passes throughequilibrium
at maximum excursion fromequilibrium    
exactly halfway between equilibrium andmaximum excursion
none of these
3 as the particle passes throughequilibrium
at maximum excursion fromequilibrium    
exactly halfway between equilibrium andmaximum excursion
none of these
3 5

Explanation / Answer

frequency f = 2.2 Hz amplitude A = 5.9 cm distance moved in 1 cycle S = 4A                                            = 23.6 cm Maximum speed = A ( 2f )   where f =frequency                           = 81.55 cm / s (2).maximum speed occur  as the particle passes throughequilibrium (3) maximum accleration at maximum excursion fromequilibrium maximum accleration = A ( 2f ) ^ 2                                  = 0.059 m * 11.27 m / s ^ 2

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