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Math Help - Free oscillations of a metre rule

  1. #1
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    Free oscillations of a metre rule

    1. The problem statement, all variables and given/known data
    Hi Everyone. I came across this problem:
    A metre rule is clamped to a table so that part of its length projects at right angles from the edge of the table. A 100g mass is attached to the free end of the rule. When the free end of rule is depressed downwards then released, the mass oscillates. Describe how you would find out if the oscillations of the mass are free oscillations.

    2. Relevant equations
    Free oscillations occur when a body oscillates with constant amplitude and frictional forces are absent.

    3. The attempt at a solution
    I proposed establishing an equilibrium position. It seems imperative that I can record vertical displacements from an equilibrium position. But I am not sure how I can measure it. At some point, air resistance should reduce the displacements. If the maximum displacement after a few measurements are not constant, the metre rule is not subject to free oscillations.
    Please comment
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  2. #2
    MHF Contributor Unknown008's Avatar
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    You can:

    Take another metre rule and place it vertically near the oscillating tip of the initial metre rule.
    Note/Mark down the reading of the vertical metre rule.
    Depress the rule and allow the mass to oscillate.
    On the vertical rule, try noting/marking the maximum and minimum heights the mass reaches as fast as possible, without touching the oscillating mass.

    Then, compare your results. If the maximum displacement upwards is equal to the maximum displacement downward, it's a free oscillation.
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  3. #3
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    Quote Originally Posted by Unknown008 View Post
    You can:

    Take another metre rule and place it vertically near the oscillating tip of the initial metre rule.
    Note/Mark down the reading of the vertical metre rule.
    Depress the rule and allow the mass to oscillate.
    On the vertical rule, try noting/marking the maximum and minimum heights the mass reaches as fast as possible, without touching the oscillating mass.

    Then, compare your results. If the maximum displacement upwards is equal to the maximum displacement downward, it's a free oscillation.
    That sounds fine. 2 questions. How can I avoid error due to parallax when taking my readings? Also, how would I position the vertical rule and ensure it is in a vertical position?
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  4. #4
    MHF Contributor Unknown008's Avatar
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    You've got to place the ruler near to the edge and stand right in front of the oscillating mass.

    As for the second, using a set square would be fine.

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    Quote Originally Posted by Unknown008 View Post
    You've got to place the ruler near to the edge and stand right in front of the oscillating mass.

    As for the second, using a set square would be fine.

    So, for the first can I presume it will be at eye level? Regarding the second, Can you expand on this technique? How does use of a set square ensure the ruler is vertical? Also, is it similar to use of a plane mirror to deduce the vertical height of a cantilever beam? Thanks
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  6. #6
    MHF Contributor Unknown008's Avatar
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    Right.

    You put the set square on horizontal ground as in the picture I gave.

    Since the ground is horizontal, the set square is exactly 90 degrees, the ruler will be at exactly 90 degrees with the ground and hence vertical.

    Um...cantilever beam?
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  7. #7
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    Quote Originally Posted by Unknown008 View Post
    Right.

    You put the set square on horizontal ground as in the picture I gave.

    Since the ground is horizontal, the set square is exactly 90 degrees, the ruler will be at exactly 90 degrees with the ground and hence vertical.

    Um...cantilever beam?
    I had a practical session where you had to obtain the vertical distance of a depressed cantilever beam. Apparently, one can do this through use of a plane mirror. I didnt know the technique then and still dont know it now. Weights will be placed on the beam and the plane mirror will be used to deduce the vertical height from its equilibrium position.
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  8. #8
    MHF Contributor Unknown008's Avatar
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    Ah I understand now! Yes, you can use that to yet reduce parallax errors, but you will have to make sure that the mirror is exactly vertical, and using the set square is just what you usually do!
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    Quote Originally Posted by Unknown008 View Post
    Ah I understand now! Yes, you can use that to yet reduce parallax errors, but you will have to make sure that the mirror is exactly vertical, and using the set square is just what you usually do!
    Cool. So, how can one use the plane mirror?
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  10. #10
    MHF Contributor Unknown008's Avatar
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    It's the same principle as with the metre rule, but you place it behind.

    Take the picture I did, the mirror will be the white background. Unfortunately I can't show you how the mirror actually reduces parallax error.

    Have you seen an analogue voltmeter or ammeter or galvanometer? Sometimes there is a small mirror along the scale. The purpose is exactly the same: to reduce parallax errors. When you look at the pointer needle, you have to make sure that the image of the needle in the mirror conincide with the image of the needle that you see. Then, you know there is no parallax error.

    The same principle is applied here.
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  11. #11
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    Quote Originally Posted by Unknown008 View Post
    It's the same principle as with the metre rule, but you place it behind.

    Take the picture I did, the mirror will be the white background. Unfortunately I can't show you how the mirror actually reduces parallax error.

    Have you seen an analogue voltmeter or ammeter or galvanometer? Sometimes there is a small mirror along the scale. The purpose is exactly the same: to reduce parallax errors. When you look at the pointer needle, you have to make sure that the image of the needle in the mirror conincide with the image of the needle that you see. Then, you know there is no parallax error.

    The same principle is applied here.

    Thanks a lot.
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