A spring of negligible mass stretches 3.00 cm from its relaxed length when a for

Question

A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 8.50 N is applied. A 0.400-kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to x = 5.00 cm and released from rest at t = 0. (Assume that the direction of the initial displacement is positive.)

(a) What is the force constant of the spring?
N/m

(b) What are the angular frequency (?), the frequency, and the period of the motion?

(c) What is the total energy of the system?
J

(d) What is the amplitude of the motion?
cm

(e) What are the maximum velocity and the maximum acceleration of the particle?

(f) Determine the displacement x of the particle from the equilibrium position at t = 0.500 s.
cm

(g) Determine the velocity and acceleration of the particle when t = 0.500 s. (Indicate the direction with the sign of your answer.)

Explanation / Answer

a)k=F/m=8.5/0.03=283.33N/m

b)w=sqrt(k/m)=26.61rad/s

f=w/2pi=4.236Hz

T=1/f=0.236s

c)U=0.5kA^2=0.5*283.33*0.05^2=0.3542J

d)A=5cm

e)vmax=Aw=1.33m/s

amax=Aw^2=35.40m/s^2

f)x=ACos(wt)=0.05Cos(26.61*0.5)=0.037m=3.7cm

g)v=-AwSin(wt)=0.896m/s

a=Aw^2Cos(wt)=26.18m/s^2

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