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September 2, 2014

September 2, 2014

Posted by **Anonymous** on Monday, November 7, 2011 at 2:49pm.

We know that

omega = 2*pi*freq = sqrt(mgL/I), where m is the mass, g is the acceleration due to gravity, and I is the moment of inertia.

So, if we can evaluate the equation above, we can get the frequency, and then take 1/freq in order to get the period.

However, don't we need the moment of Intertia, I, for the meter stick?

How do we solve this?

Thanks!

- Physics -
**drwls**, Monday, November 7, 2011 at 2:59pmThey are both meter sticks. Note that M/I appears in the frequency equation. I is proportional to M so Mass cancels out.

They will both have the same period.

If they swing from one end, I = M L^2/3

M/I = L^2/3

- Physics -
**Anonymous**, Monday, November 7, 2011 at 3:27pmMaybe I don't understand..

If I=m (L^2) /3, then we have:

w=2 pi f = sqrt(mgL/I)

Subbing in I, we have:

2 pi f = sqrt(g L / (L^2/3) )

=sqrt(3g/L)

Thus, we have:

f=1/(2pi) * sqrt(3*g/L)=1/(2pi) * sqrt(3*9.8/1) = 0.862992.

This would mean that the period should be:

1/ freq = 1/0.862992, but I don't believe this answer is correct.

Can you please clarify?

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