High School

A performer seated on a trapeze is swinging back and forth with a period of 8.85 s. 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 the trapeze-performer system as a simple pendulum.

Answer :

If she stands up, thus raising the center of mass of the trapeze + performer system by 35.0 cm, the new period of the system will be 8.79 s.

Given, T = 8.85 s

ΔL = 35.0 cm = 0.35 m

Here, we will apply the formula of time period of simple harmonic motion of pendulum.

T = 2π [tex]\sqrt{L/g[/tex]

8.85 = 2π [tex]\sqrt{L/9.8}[/tex]

L = 19.46 m

After standing up new length

L' = L - ΔL

L' = 19.46 - 0.35

L' = 19.11 m

So, new time period will be

T' = 2π [tex]\sqrt{L'/g}[/tex]

T' = 2π [tex]\sqrt{19.11/9.8}[/tex]

T' = 2 × 3.14 × 1.3964

T' = 8.76 s

So, The new time period will be 8.76 s.

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