Answer :
Final answer:
To find the new period of the trapeze + performer system, we need to calculate the change in length caused by the performer standing up and use the equation for the period of a simple pendulum.
Explanation:
To solve this problem, we need to recognize that the swinging of the performer on the trapeze can be modeled as a simple pendulum. The period of a simple pendulum can be calculated using the equation T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
In this case, when the performer stands up, the center of mass of the trapeze + performer system is raised by 44.0 cm. This change in the length of the pendulum will affect its period.
We can calculate the new period using the equation T' = 2π√((L+d)/g), where T' is the new period, L is the original length of the pendulum, d is the change in length, and g is the acceleration due to gravity.
Plugging in the values, we get T' = 2π√((L+0.44)/g). This equation gives us the new period of the system.
Answer:
8.800s
Explanation:
When the performer swings, she oscillates in SHM about Lo of the string with time period To = 8.90s.
First, determine the original length Lo, where for a SHM the time period is related to length and the gravitational acceleration by the equation
T = 2π×√(Lo/g)..... (1)
Let's make Lo the subject of the formulae
Lo = gTo^2/4π^2 ..... (2)
Let's put our values into equation (2) to get Lo
Lo = gTo^2/4π^2
= (9.8m/s^2)(8.90s)^2
------------------------------
4π^2
= 19.663m
Second instant, when she rise by 44.0cm, so the length Lo will be reduced by 44.0cm and the final length will be
L = Lo - (0.44m)
= 19.663m - 0.44m
= 19.223m
Now let use the value of L into equation (1) to get the period T after raising
T = 2π×√(L/g)
= 2π×√(19.223m/9.8m/s^2)
= 8.800s
