Question

An infant's toy has a 120 g wooden animal hanging from a spring. If pulled down gently, the animal oscillates up and down with a period of 0.47s . His older sister pulls the spring a bit more than intended. She pulls the animal 27cm below its equilibrium position, then lets go. The animal flies upward and detaches from the spring right at the animal's equilibrium position.

If the animal does not hit anything on the way up, how far above its equilibrium position will it go?

Answer 1

The key here is observing that since the spring detaches right at the equilibrium position, the energy in the system at that point is entirely kinetic. We can use conservation of energy to find the velocity and then some kinematic equations to find the maximum height. First, we'll need the spring constant.

The period of a spring oscillator is
T = 2pi * sqrt(m/k)
T^2 = 4pi^2 * m/k
k = 4pi^2 * m/T^2 = 4pi^2 * .120 kg / (0.47 sec)^2 = 21.424 N/m

Now the potential energy in the spring when the sister lets go equals the kinetic energy at detachment:

U = KE
1/2 * kx^2 = 1/2 * mv^2
v^2 = kx^2 / m
v = x * sqrt(k/m) = 0.27 m * sqrt (21.424 N/m / 0.120 kg) = 3.607 m/s

And now the height, when velocity = 0...
v^2 = vo^2 + 2as
0 = vo^2 -2gs
s = vo^2 / 2g = (3.607 m/s)^2 / (2*9.8 m/s^2) = 0.664 m