Question

Two particles of masses m1 and m2 are joined by a massless spring of natural length L and force constant k. Initially, m2 is resting on a table and m1 is held vertically above m2 at a height L at time t = 0, m1 is projected vertically upward with initial velocity v0. Find the positions of the two masses at any subsequent time t ( before either mass returns to the table) and describe the motion.

Answer 1

When m1 is projected vertically upward, it undergoes simple harmonic motion (SHM) and oscillates between its maximum height and the equilibrium position. The position of m2 remains at rest on the table.

Step-by-step explanation:

To determine the positions of the two masses at any subsequent time t, we can apply the principles of simple harmonic motion (SHM) and conservation of energy. As m1 is projected vertically upward, it will undergo SHM and oscillate between its maximum height and the equilibrium position, which is at the natural length L of the spring. The position of m2, on the other hand, will remain on the table as it is at rest.

Let's consider the position of m1. The equation of motion for m1 can be given as:

x1(t) = L + A1 * cos(ωt + φ1)

Where:

• x1(t) is the position of m1 at time t.
• L is the natural length of the spring.
• A1 is the amplitude of m1's oscillation (maximum distance from the equilibrium position).
• ω is the angular frequency of the oscillation.
• φ1 is the phase constant, representing the initial condition of the oscillation.

As for m2, it will remain at rest on the table, so its position remains constant at x2 = 0.

In summary, at any subsequent time t, m1 will oscillate between its maximum height (L+A1) and the equilibrium position (L), while m2 will remain at rest on the table (x2 = 0).