Question

A 2.9-kg cart is attached to a horizontal spring for which the spring constant is 60 N/m . The system is set in motion when the cart is 0.27 m from its equilibrium position, and the initial velocity is 1.8 m/s directed away from the equilibrium position.

What is the amplitude of the oscillation?

What is the speed of the cart at its equilibrium position?

Answer 1

Mass of the cart = m = 2.9 kg

Spring constant = k = 60 N/m

Initial displacement = X₁ = 0.27 m

Initial velocity = V₁ = 1.8 m/s

Amplitude of the oscillation = A

Speed of the cart at equilibrium position = V

Total mechanical energy of the system is equal to the sum of the potential energy of the spring and the kinetic energy of the cart.

Total mechanical energy at maximum displacement is equal to the potential energy of the spring.

kA²/2 = kX₁²/2 + mV₁²/2

kA² = kX₁² + mV₁²

(60)A² = (60)(0.27)² + (2.9)(1.8)²

A = 0.479 m

At the equilibrium point the total mechanical energy is converted into kinetic energy of the cart.

kA²/2 = mV²/2

kA² = mV²

(60)(0.479)² = (2.9)V²

V = 2.179 m/s

a) Amplitude of oscillation = 0.479 m

b) Speed of the cart at equilibrium position = 2.179 m/s