A performer, seated on a trapeze, is swinging back and forth with a period of 8.90 s. If she stands up, thus raising the center of mass of the trapeze + performer system by 44.0 cm, what will be the new period of the system? Treat trapeze + performer as a simple pendulum.

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

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straight or amphibians are homeothermic to thermal