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 8.00 N is applied. A 0.470-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?
1 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?
5 J

(d) What is the amplitude of the motion?
6 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.
9 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) F = k*x
==> k = F/x = 8/0.03 = 266.67 N/m

b)

angular frequency, w = sqrt(k/m) = sqrt(266.7/0.47) = 23.82 rad/s

frequency, f = w/2*pi = 23.82/(2*3.14) = 3.79 Hz

Time periode, T = 1/f = 0.264 s

c) U = 0.5*k*x^2 = 0.5*266.67*0.05^2 = 0.333 J

d) A = 5 cm

e)

0.5*m*Vmax^2 = U

Vmax = sqrt(2*U/m) = sqrt(2*0.333/0.47) = 1.19 m/s

Fmax = k*A

m*amax = k*A

amx = k*A/m = 266.67*0.05/0.47 = 28.37 m/s^2

f)

x = A*cos(w*t)

= 0.05*cos(23.82*0.5)

= 0.0396

= 3.96 cm

g)

v = -A*w*sin(w*t)

= -0.05*23.82*sin(23.82*0.5)

= 0.7126 m/s

a = -A*w^2*cos(w*t)

= -0.05*23.82^2*cos(23.82*0.5)

= -22.47 m/s^2

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