Two equal masses are attached to separate identical springs next to one another. One mass is pulled so its spring stretches 20 cm and the other is pulled so its spring stretches only 10 cm. The masses are released simultaneously.Which mass reaches the equilibrium point first?

Why would they do it in different times? Why should amplitude affect frequency?

To determine which mass reaches the equilibrium point first, we need to understand the relationship between the period of oscillation and the stiffness of the spring.

The period of oscillation, T, is the time taken for one complete cycle of motion. It can be calculated using the formula:

T = 2π √(m/k)

Where m is the mass and k is the spring constant.

In this case, we have two identical masses, so the only difference is the spring constant. Let's compare the two springs.

The spring constant, k, is a measure of how stiff the spring is. A higher value of k indicates a stiffer spring.

Given that one spring stretches 20 cm and the other stretches 10 cm, we can assume that the spring with the 20 cm stretch has a higher spring constant (k1) than the one with the 10 cm stretch (k2). This is because a stiffer spring would stretch less for the same mass.

Since we have T = 2π √(m/k), we can reason that the mass with the lower spring constant will have a longer period of oscillation.

Therefore, the mass attached to the spring with the 10 cm stretch will reach the equilibrium point first.