Physical Pendulum oscillation problem

[sapiental]

(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 I am doing wrong?

Thanks

 

[OlderDan]

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.

 

[kyle8921]

for your question. It seems that you have made a minor calculation error in your solution. The correct answer should be (a) 0.67 Hz. Here is the correct calculation:

The moment of inertia of the meter stick about its center of mass is mL^2/12, where m is the mass of the stick and L is its length. In this case, m = 1 kg and L = 1 m, so the moment of inertia is 1/12 kg*m^2.

The distance from the pivot point to the center of mass is 80 cm, or 0.8 m. This means that the distance from the pivot point to the end of the stick is 0.2 m.

Using the formula T = 2pi*sqrt(I/(mgd)), we get T = 2pi*sqrt((1/12)/(1*9.8*0.2)) = 2pi*sqrt(5/12) = 2pi*0.645 = 4.05 seconds.

The frequency is then f = 1/T = 1/4.05 = 0.247 Hz = 0.67 Hz (rounded to two decimal places).

I hope this helps clarify any confusion and shows you where your calculation went wrong. Keep up the good work!