High School

A performer seated on a trapeze is swinging back and forth with a period of 9.21 s. If she stands up, raising the center of mass of the trapeze-performer system by 40.9 cm, what will be the new period of the system?

Answer :

the new period of the system will be longer than the original period of 9.21 s.

we need to consider the relationship between the period of a pendulum and its length. The period of a pendulum is directly proportional to the square root of its length. When the performer stands up, the center of mass of the system is raised, increasing the length of the pendulum. This means that the period of the system will also increase.

We can use the formula for the period of a pendulum, T = 2π√(L/g), where T is the period, L is the length, and g is the acceleration due to gravity. We can rearrange this formula to solve for the new period:

T₂ = 2π√(L₂/g)

where L₂ is the new length of the pendulum after the performer stands up.

To find L₂, we need to add the distance that the center of mass was raised to the original length of the pendulum.

L₂ = L₁ + d

where L₁ is the original length of the pendulum and d is the distance that the center of mass was raised.

Substituting this into the formula for the new period, we get:

T₂ = 2π√((L₁ + d)/g)

We can now plug in the values given in the problem:

T₂ = 2π√((L₁ + 0.409 m)/9.81 m/s²)

T₂ = 2π√((L₁ + 0.0417)/1.00)

T₂ = 2π√(L₁ + 0.0417)

We don't have a numerical value for L₁, but we can see that the new period, T₂, will be longer than the original period, T₁.

when the performer stands up on the trapeze, the new period of the system will be longer than the original period of 9.21 s. This is because the center of mass of the system is raised, increasing the length of the pendulum and therefore its period.

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