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 0.540-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.)

? = rad/s f = Hz T = s
(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? vmax = m/s amax = m/s2
(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.)
v = m/s a = m/s2

Explanation / Answer

a) F = k*x
==> k = F/x = 8/0.03 = 266.67 N/m

b)

angular frequency, w = sqrt(k/m) = sqrt(266.7/0.54) = 22.22 rad/s

frequency, f = w/2*pi = 22.22/(2*3.14) = 3.54 Hz

Time periode, T = 1/f = 0.283 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*0.333/0.54) = 1.11 m/s

Fmax = k*A

m*amax = k*A

amx = k*A/m = 266.67*0.05/0.54 = 24.7 m/s^2

f)

x = A*cos(w*t)

= 0.05*cos(22.22*0.5)

= 0.0057

= 0.57 cm

g)

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

= -0.05*22.22*sin(22.22*0.5)

= 1.104 m/s

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

= -0.05*22.22^2*cos(22.22*0.5)

= -2.82 m/s^2

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