Question: A performer seated on a trapeze is swinging back and forth with a period of 8.85s . If she stands up, thus raising the center of mass of the trapeze+ performer system by 35.0 cm , what will be the new period of the system? Treat trapeze+ performer as a simple pendulum.

1. The period of swinging of the trapeze is T = 8.85 s
2. For standing, the center of mass of the system is $h=35.0\text{cm or}35.0\times {10}^{-2}\text{m}$

The simple pendulum oscillates to perform a simple harmonic motion. In simple harmonic motion, the acceleration is proportional to the displacement of the particle and is always directed towards the mean position. The time period of oscillation of a pendulum can be defined in terms of its length and gravitational acceleration.

Formulae:

$T=2\pi \sqrt{\frac{L}{g}}$

Here, T is the time period of oscillation, L is the length of pendulum, g is gravitational acceleration.

Consider the system (trapeze + performer) is a simple pendulum.The period of oscillation for the simple pendulum is,

$T=2\pi \sqrt{\frac{L}{g}}$

When the performer is sitting, the length of the system is

$$\begin{gathered} T^2=4{\pi }^2\left(\frac{L}{g}\right) \\ L=\frac{T^{2}g}{4{\pi }^2} \end{gathered}$$

When the performer is standing, the new length of the system is

$L\text{'}=L-h$

The new period of oscillation for the simple pendulum is

$$\begin{gathered} T\text{'}=2\pi \sqrt{\frac{L\text{'}}{g}} \\ =2\pi \sqrt{\frac{L-h}{g}} \\ =2\pi \sqrt{\frac{\left(\frac{T^{2}g}{4{\pi }^2}\right)-h}{g}} \\ =2\pi \sqrt{\frac{\left(\frac{{\left(8.85\text{s}\right)}^2\times 9.8{\text{m/s}}^2}{4\times {\left(3.14\right)}^2}\right)-35.0\times {10}^{-2}\text{m}}{9.8{\text{m/s}}^2}} \\ =8.77\text{s} \end{gathered}$$

Therefore, the period of oscillation of the system as (trapeze + performer) for standing is 8.77s .