High School

A performer seated on a trapeze is swinging back and forth with a period of 9.29 seconds. If she stands up, thus raising the center of mass of the trapeze and performer system by 38.7 cm, what will be the new period of the system? Treat the trapeze and performer as a simple pendulum.

Answer :

The new period of the system, when the performer stands up, is approximately 2.487 seconds.

To solve this problem, we can use the formula for the period of a simple pendulum:

T = 2π√(L/g)

where:

T is the period of the pendulum,

L is the length of the pendulum, and

g is the acceleration due to gravity.

Initially, the performer is seated on the trapeze, and we need to find the initial length of the pendulum. Let's denote it as L_initial.

Given:

Period of the system when seated (T_initial) = 9.29 s

Rearranging the formula, we can solve for L_initial:

L_initial = (T_initial/2π)^2 * g

Now, let's calculate the new length of the pendulum when the performer stands up. We can use the concept of the center of mass to determine this.

Given:

Change in height (Δh) = 38.7 cm = 0.387 m

The new length of the pendulum (L_new) will be the sum of the initial length (L_initial) and the change in height (Δh).

L_new = L_initial + Δh

Finally, we can calculate the new period (T_new) using the formula for the period of a simple pendulum with the new length:

T_new = 2π√(L_new/g)

Let's plug in the given values and calculate the new period:

Using g = 9.8 m/s² (acceleration due to gravity), we have:

L_initial = (9.29/2π)² * 9.8 = 1.152 m

L_new = 1.152 + 0.387 = 1.539 m

T_new = 2π√(1.539/9.8) ≈ 2π√0.157 ≈ 2π * 0.396 ≈ 2.487 s

Therefore, the new period of the system, when the performer stands up, is approximately 2.487 seconds.

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