A particle that hangs from a spring oscillates with an angular frequency of 2 rad/s. The spring is suspended from the ceiling of an elevator car and hangs motionless (relative to the car) as the car descends at a constant speed of 1.5 m/s. The car then suddenly stops. Neglect the mass of the spring. (a) With what amplitude does the particle oscillate? (b) What is the equation of motion for the particle? (Choose the upward direction to be positive.) Solution: (a) When traveling in the elevator at constant speed, the total force on the mass is zero. The force exerted by the spring is equal in magnitude to the gravitational force on the mass, the spring has the equilibrium length of a vertical spring. When the elevator suddenly stops, the end of the spring attached to the ceiling stops. The mass, however has momentum, p = mv, and therefore starts stretching the spring. 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). Since vmax = ωA and ω = 2/s, the amplitude of the amplitude of the oscillations is A = 0.75 m. (b) The equation of motion for the particle is d2x/dt2 = -(k/m)x = -ω2x. Its solution is x(t) = Acos(ωt + φ) = (0.75 m)cos((2/s)t + φ). If we choose the t = 0 to be the time the elevator stops and let the upward direction be positive, then x(0) = 0, and v(0) = -1.5 m/s. We therefore need φ to be π/2. Issues: Can someone explain that when the elevator hasn't stopped, is the block executing SHM inside the elevator? If yes, then how can the force equal its weight when both of them are pointing downwards (that would be when the block is going upwards). If not, then what force are we talking about here?