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 7.80 N is applied. A 0.480-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

To find G), we need to make sure B is correct first (what you got for rad/s):

? =?(k/m) = ?(260/0.48)

= 23.27 rad/s [ANS]

Now for G, we know that  v = -A? Sin(?t)

so:  v = - ( 5.00*10^-2 m ) (23.27 rad /s) sin ( 23.27 * 0.500s) = - (-0.934) = 0.934 m/s [ANS]

And for acceleration:

a = A ?^2 cos ?t = - ( 5.00*10^-2 m ) ( 23.27 rad/s)^2 cos ( 23.27 * 0.500) = 16.16 m/s^2 [ANS]

Hope this helps :D

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