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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.

Short Answer

Expert verified

Answer

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

Step by step solution

01

Identification of given data 

  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=35.0cm or35.0ร—10-2m
02

Understanding the concept

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=2ฯ€Lg

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

03

Determining the new period of the system

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

T=2ฯ€Lg

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

T2=4ฯ€2LgL=T2g4ฯ€2

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

L'=L-h

The new period of oscillation for the simple pendulum is

T'=2ฯ€L'g=2ฯ€L-hg=2ฯ€T2g4ฯ€2-hg=2ฯ€8.85s2ร—9.8m/s24ร—3.142-35.0ร—10-2m9.8m/s2=8.77s

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

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