A vertically Hanging Spring Loaded with a Mass: The upper end of a massless spri
ID: 1507568 • Letter: A
Question
A vertically Hanging Spring Loaded with a Mass: The upper end of a massless spring of constant k is attached to a fixed point. A box of mass m is hung at the lower end of the spring. Initially, you hold this box at rest with your hand at a vertical position so that the spring is neither stretched nor compressed. For all the processes described below, the spring remains completely elastic. Please set up a vertical y-axis pointing up with this resting position as the origin. If you lower your hand slowly, the box will follow your hand initially. At some vertical position, the box will stop and no longer follow your hand. This is the equilibrium position. Find this equilibrium position. Now, again, you hold this box at that initial resting position so that the spring is neither stretched nor compressed. This time, instead of lowering the box slowly as in (a), you are going to release the box at time t = 0 and let it drop "freely". Find the time taken by the box to return to your hand.Explanation / Answer
Use conservation of energy. Consider the gravitational potential energy reference to be the point where it will eventually quit bouncing, the equilibrium point which is d below the starting point. Then the box potential just before it is dropped is
GPE = m*g*d
When the downward motion stops, all of the original GPE will be in the form of spring potential energy.
SPE = (1/2)*k*y^2
where y is measured from the equilibrium point and k is the spring's force constant given by
F = k*y.