A spring-mounted aluminum ball oscillates with a period of 2 s.

A spring-mounted aluminum ball oscillates with a period of 2 s. What will be the period of oscillation if the aluminum ball is replaced with a copper of the same volume?

Ta = 2 s.

ρа = 2700 kg / m3.

ρm = 8920 kg / m3.

Vа = Vm.

Tm -?

The period of oscillation of a spring pendulum T is the time of one complete oscillation. For a spring pendulum, the period of free natural oscillations T is determined by the formula: T = 2 * P * √m / √k, where P is the number pi, m is the mass of the load attached to the spring, k is the spring stiffness.

We express the mass of the cargo m by the formula: m = V * ρ, where V is the volume of the body, ρ is the density of the material from which the cargo is made.

Let us express the period of free natural vibrations of the aluminum and copper load: Ta = 2 * P * √ (V * ρa) / √k, Tm = 2 * P * √ (V * ρm) / √k.

2 * P / √k = Ta / √ (V * ρа).

Tm = Ta * √ (V * ρm) / / √ (V * ρa) = Ta * √ρm / √ρa.

Tm = 2 s * √8920 kg / m3 / √2700 kg / m3 = 3.63 s.

Answer: the period of free natural oscillations of the copper load is Tm = 3.63 s.