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 8.70 N is applied. A 0.550-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.)

(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

let k is the spring constant of the spring.

Apply, F = k*x

k = F/x

= 8.7/0.03

= 290 N/m

angualr frequency of the motion when the particle is attached with the spring, w = sqrt(k/m)

= sqrt(290/0.55)

= 22.96 rad/s

Amplitude, A = 5 cm

so we can write equation of motion, x = A*cos(w*t)

x = 5*cos(22.96*t)

at t = 0.5 s

x = 5*cos(22.96*0.5)

= 2.33 cm <<<<<<<---------Answer

g) x = dv/dt

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

= -5*22.96*sin(22.96*0.5)

= 102 cm/s

= 1.02 m/s <<<<<<<---------Answer

a = dv/dt

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

= -w^2*x

= -22.96^2*0.0233

= -12.3 m/s^2 <<<<<<<---------Answer

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