Question

A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.30 N is applied. A 0.520-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. Use the exact values you enter to make later calculations.)

(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

Answer 1

7.3 N force stretches 3 cm

so,

force = kx

7.3 = k * 0.03

k = 243.33 N/m

angular frequency = sqrt(k / m)

angular frequency = sqrt(243.33 / 0.52)

angular frequency = 21.63 rad/sec

equation of particle displacement

y(t) = A * cos(wt)

A = 5 cm

at time t = 0.5 sec

y(0.5) = 0.05 * cos(21.63 * 0.5)

f) y(0.5) = -0.00897 m or -0.897 cm

velocity(t) = -Aw * sin(wt)

velocity(0.5) = -0.05 * 21.63 * sin(21.63 * 0.5)

g) velocity(0.5) = 1.0639 m/s

acceleration(t) = Aw^2 * cos(wt)

acceleration(0.5) = 0.05 * 21.63^2 * cos(21.63 * 0.5)

g) acceleration(0.5) = -4.201 m/s^2