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

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the second one

To determine which mass reaches the equilibrium point first, we need to consider the spring constants of the two springs. The spring constant is a measure of the stiffness of the spring and is denoted by the symbol k. The equation for the period of oscillation T of a mass-spring system is given by:

T = 2π√(m/k)

where m is the mass attached to the spring.

Since the masses are equal, the mass m cancels out, and the equation simplifies to:

T = 2π√(1/k)

To compare the periods of oscillation, we need to compare the square roots of the spring constants.

Let's assume the spring constant of the first spring is k1, and the spring constant of the second spring is k2.

Given that the first spring stretches 20 cm and the second spring stretches 10 cm, we can relate the two spring constants using the equation for the spring force, F = kx, where F is the force applied to the spring and x is the displacement (stretch or compression) of the spring.

For the first spring:
F1 = k1 * (20 cm)

For the second spring:
F2 = k2 * (10 cm)

Since the masses are equal, the magnitudes of the forces F1 and F2 must also be equal for both masses. Therefore, we have:

F1 = F2

k1 * (20 cm) = k2 * (10 cm)

Simplifying this equation, we find:

k1 = 2 * k2

This means the spring constant of the first spring is twice the spring constant of the second spring.

Now, let's substitute this relationship into the equation for the period of oscillation:

T1 = 2π√(1/k1)

T2 = 2π√(1/k2)

Since k1 = 2 * k2, we have:

T1 = 2π√(1/(2 * k2))

T2 = 2π√(1/k2)

Comparing the expressions for T1 and T2, we can see that T1 is smaller than T2 because the square root of 1/(2 * k2) is smaller than the square root of 1/k2.

Therefore, the mass attached to the first spring, which has a larger spring constant, will reach the equilibrium point first.