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 8.90 N is applied. A 0.510-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.)

b) What are the angular frequency , the frequency, and the period of the motion?

c) What is the total energy of the system?

d) What is the amplitude of the motion?

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.

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

fro spring - mass :

F = k x

8.90 = k (0.03)

k = 296.67 N/m

b) m w^2 = k

w = sqrt(k/m) = sqrt(296.67 / 0.510) = 24.12 rad/s ........Ans

f = w / 2pi = 3.84 hz .........Ans

T = 1/f = 0.26 sec .........Ans

c) total energy = k A^2 /2

        = 296.67 ( 0.05^2) / 2 =0.37 J

d) A = 0.05 m

e)
Vmax = Aw = 24.12 x 0.05 = 1.21 m/s

a_max = A w^2 = 24.12^2 x 0.05 =29.1 m/s^2

f) x = A cos(wt)

x = 0.05 m cos(24.12x 0.5) = 0.0437 m   Or 4.37 cm

g) v = - A w sin(wt) = - 1.21 sin(24.12 x 0.5) = 0.59 m/s

a_max = - A w^2 cos(wt) = - 29.1 cos(14.12 x 0.5) = - 25.45 m/s^2

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