Question

The displacement as a function of time of a 4.0kg mass spring simple harmonic oscillator is .  

What is the displacement of the mass at 2.2 seconds? ___________m

What is the spring constant? ___________________N/m

What is the position of the object when the speed is maximum? ______________m

What is the magnitude of the maximum velocity?____________________m/s

Answer 1

Answer:

You have missed the expression for the displacement x(t), but I knew this problem that is why I am using the displacement's expression as far as my knowledge.

x(t) = 20.0m cos(6t)

From this expression, amplitude is A = 20.0 m and angular frequency is = 6 = 18.84 rad/s

Mass m = 4.0 kg

(1) At time t = 2.2 s, the displacement of the mass is x(t = 2.2 s) = 20.0m cos[6(2.2 s)] = -16.42 m.

(2) Spring constant k = m² = (4.0 kg) (6)² = 1419.78 N/m.

(3) The mass will reach to its maximum speed at the equilibrium position, that means at x = 0 m.

At the equilibrium position the potential energy U stored in the spring is totally converted into kinetic energy KE. That is why at x = 0 m, the mass has maximum speed.

(4) The magnitude of maximum speed of the mass is v_max = A = (18.84 rad/s) (20.0 m) = 376.8 m/s.