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 7.00 N is applied. A 0.450-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) F = kx

k = F/x = 7/0.03 = 233 N/m

b) w = sqrt(k/m ) = sqrt(233/0.45) = 22.75 rad/s

T = 2*3.1415/22.75 = 0.276 s

f = 1/ t = 1/0.276 = 3.62 Hz

c) E = 0+ 0.5*233*0.05^2 = 0.29 J

d) E = 0.5 kA^2

A = sqrt(2E/K ) = sqrt(2*0.29/233) = 0.05 m

e) V_max = sqrt(2*E/m) = 1.135 m/s

Q_max = F_max/, = KA/m = 233*0.05/0.45 = 25.88 m/s^2

f) x = 0.05(cos[0.5(srt(233/0.45))) = 0.049 m

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