Question

Две бусинки массами m и 2m надеты на две параллельные горизонтальные спицы, находящиеся на расстоянии L друг от друга. Бусинки связаны резинкой, натяжение которой равно F. Найти период колебаний бусинок, возникающих при их небольшом смещении от положения равновесия. Трение между бусинками и спицами отсутствует.

Answer 1

Неподвижная точка резинки делит ее пропорционально массам бусинок. Значит она находится на расстоянии 1/3 L от большей массы.
Жёсткость резинки , которая ближе к большей массе равна 3F/L
Период колебаний равен 2π√(m/k)
T = 2π√(2mL/3F)

Проверим расчет для меньшей массы
Жёсткость 3F/2L
Период
Т2=2π√(2mL/3F)

Answer 2

Problem Analysis

We are given two beads with masses m and 2m, respectively, connected by a rubber band. The beads are threaded onto two parallel horizontal spokes that are a distance L apart. The rubber band is under tension F. We need to find the period of oscillation for the beads when they are displaced slightly from their equilibrium position. Friction between the beads and the spokes is assumed to be absent.

Solution

To find the period of oscillation, we can use the formula for the period of a simple harmonic oscillator:

T = 2π√(m_eff/k_eff)

where T is the period, m_eff is the effective mass, and k_eff is the effective spring constant.

Finding the Effective Mass

The effective mass of the system can be found by considering the motion of the two beads as a single system. Since the beads are connected by a rubber band, they will move together. The total mass of the system is m + 2m = 3m.

Finding the Effective Spring Constant

The rubber band provides the restoring force for the oscillation. The tension in the rubber band, F, is equal to the force required to stretch or compress the rubber band. This force is proportional to the displacement of the beads from their equilibrium position.

The force required to stretch or compress the rubber band can be modeled as a spring force. The spring constant, k, is given by:

k = F/x

where x is the displacement of the beads from their equilibrium position.

Since the beads are connected by the rubber band, they will move together. The displacement of the beads from their equilibrium position is the same for both beads. Therefore, we can write:

k_eff = k/2

Substituting Values and Calculating the Period

Substituting the values into the formula for the period of oscillation, we have:

T = 2π√(m_eff/k_eff) = 2π√((3m)/(k/2)) = 2π√((6m)/k)

Final Answer

The period of oscillation for the beads, when displaced slightly from their equilibrium position, is 2π√((6m)/k).

Please note that this solution assumes that the beads are small compared to the distance between the spokes and that the motion is approximately simple harmonic.