A 0.20 kg object, attached to a spring with spring constant k = 10 N/m, is moving on a horizontal frictionless surface in simple harmonic motion of amplitude of 0.080 m. What is its speed at the instant when its displacement is 0.040 m? (Hint: Use conservation of energy.)

Answer 1

The speed of the object at the instant when its displacement is 0.040 m can be calculated using the conservation of energy equation.

Speed = sqrt(2*(0.5*k*(0.080 m)^2 - 0.5*k*(0.040 m)^2))/0.20 kg

Speed = sqrt(2*(0.8 - 0.2))/0.20 kg

Speed = sqrt(0.6)/0.20 kg

Speed = 1.2 m/s