Question

if a mass of 200g is attached to a spring with spring constant 0.8m and pulled by 5 cm then potential energy at equilibrium position (x=0) is​

Answer 1

Answer:

It moves through the equilibrium position of the vertical spring with its maximum velocity vmax = 1.5 m/s. Its velocity as a function of time is v(t) = -ωAsin(ωt + φ). Details of the calculation: Since vmax = ωA and ω = 2/s, the amplitude of the amplitude of the oscillations is A = 0.75 m.