Time period of oscillation of bar magnet

  • Thread starter Amith2006
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  • #1
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1)A thin rectangular magnet suspended freely has a period of oscillation of 4 seconds. If it is broken into 2 halves (each having half the original length) and one of the pieces is suspended similarly. What is the new period of oscillation?
I solved it in the following way:
Let E1 and E2 be the moment of inertia of the magnets of length L and L/2 respectively.
E1 = M(L^2)/12
E2 = (M/2)(L^2/4)/12 = E1/8
i.e. (E1/E2) = 8
Here M is the mass of the magnet.
Let T1 and T2 be the initial and final time period.
T is proportional to (E)^(1/2)
Here m is the magnetic dipole moment of the magnet. The dipole moment of the magnet doesn’t change because the magnet is cut along the perpendicular bisector of its axis.
(T1/T2) = (E1/E2)^(1/2)
(4/T2) = (8)^(1/2)
Solving I get,
T2 = sqrt(2) seconds
But the answer given in my book is 2 seconds. Please guide me.
 

Answers and Replies

  • #2
Doc Al
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Amith2006 said:
Let T1 and T2 be the initial and final time period.
T is proportional to (E)^(1/2)
This is a physical pendulum and must be treated as such. The period of a physical pendulum is given by:
[tex]T = \sqrt{\frac{I}{Mg L_{cm}}}[/tex]

I = rotational inertia about the pivot point (which I assume to be the end of the magnet, not its center);
[itex]L_{cm}[/itex] = the distance from the pivot point to the center of mass.
 
  • #3
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That's cool! Thanks.
 

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