Notices
Results 1 to 4 of 4

Thread: period of oscillation of bar magnet

  1. #1 period of oscillation of bar magnet 
    Forum Freshman
    Join Date
    Jun 2006
    Posts
    26
    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 its perpendicular bisector.
    (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.


    Reply With Quote  
     

  2.  
     

  3. #2 Re: period of oscillation of bar magnet 
    Forum Ph.D. william's Avatar
    Join Date
    Jun 2006
    Location
    Wherever I go, there I am
    Posts
    935
    Quote Originally Posted by Photon
    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 its perpendicular bisector.
    (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.
    Hi photon,
    I'm not sure I'm picturing it correctly (a bar magnet rotating?) but I think you are correct. I get the same answer as you from energy arguments - plus, your math seems right. (Plus "2 seconds" seems like too nice of a number....)

    Are you sure the book is correct?
    If so, when you see how the problem is done to get 2 seconds, you'll have to show us.

    Cheers,
    william


    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
    Reply With Quote  
     

  4. #3  
    M
    M is offline
    Forum Junior
    Join Date
    Sep 2006
    Posts
    282
    The problem is ill-posed.

    a) What type of "oscillation" are you talking about? (There is more than one possibility). Since you use the moment of inertia, it should be an oscillatory motion of some sort (unlike, for example, an internal structural vibration).

    b) The magnetic dipole moment does not enter your equations in any way. How does it matter that this is a magnet? Don't you think there should be some significance to that quality?

    I suspect the answers to a) and b) will be related and shed some light on the problem. Think about it and try to properly pose the problem and you might actually come up with the solution yourself.
    Reply With Quote  
     

  5. #4  
    M
    M is offline
    Forum Junior
    Join Date
    Sep 2006
    Posts
    282
    scratch question b). It may just be a magnet to justify the "free suspension".

    You still need to say how the magnet is moving, though. Your choice of moment of intertia implies rotation about a particular axis. Could it be you are considering the wrong axis?

    Essentially, this looks a lot like a momentum conservation problem. The angular momentum of the new system (2 halfs) has to be the same as the old (1 piece). Angular momentum is moment of inertia times angular velocity. You take it from here.

    You say that only "ONE of the pieces is suspended similarly", which also seems odd. Is that the original text from the book? Does that mean all momentum goes to one half while the other half falls down?
    Reply With Quote  
     

Bookmarks
Bookmarks
Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •