Two physical pendulums (not simple pendulums) are made from meter sticks that are suspended from the ceiling at one end. The sticks are uniform and are identical in all respects, except that one is made of wood (mass=0.34 kg) and the other of metal (mass=0.63 kg) They are set into oscillation and execute simple harmonic motion. Determine the period of (a) the wood pendulum and (b) the metal pendulum

Question

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!

Answer 1

They 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

Answer 2

Maybe 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?