Physical Pendulum oscillation problem

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(9) A physical pendulum consists of a meter stick (1 meter long) pivoted at a distance 20 cm
from one end and suspended freely. The frequency for small oscillation is closest to
(a) 0.67 Hz (b) 0.8 Hz (c) 1.1 Hz (d) 1.7 Hz (e) Insufficient information
(Hint: The moment of inertia of a stick of mass m and length L about its center of mass is mL2/12.)

Professor says the correct answer is (a) 0.67 Hz

I don't know what I do wrong on this question but I always get the frequency equal to .78Hz which is closer to .8Hz.

T = 2pi sqrt(I/(mgd)) I = moment of inertia

so T = 2pi sqrt((1/12)mL^2) / (mgL(1/5)))

T = 2pi sqrt(5L/12g)

T = 2pi sqrt(5/(12 * 9.8))

T = 1.29

f = 1/1.29 = .775

can someone tell me what im doing wrong?

Thanks
 

Answers and Replies

  • #2
OlderDan
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The moment of inertia you are using is wrong. The stick is not pivoting about its center. If it were there would be no oscillations at all. You will need to use the parallel axis theorem.
 

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