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Math Help - more questions on simple harmonic motion

  1. #1
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    more questions on simple harmonic motion

    Could someone check these for me? i was told to put g = 9.8 m/s for all calculus questions, and I think it's a bit ridiculous.

    1. A long jumper running at 8 m/s jumps into the air, rising to a height of 0.7 metres. What is the length of the jump.

    maximum height
     0.7 = \frac{V^2sin^2\alpha }{2g}

     0.7 = \frac{8^2sin^2\alpha }{2\cdot 9.8}

     13.72 = 64sin^2\alpha

     sin\alpha = 0.463006479

    \alpha = 27^{\circ}{35}'

    Range
     \frac{V^2sin2\alpha }{9.8}

     =\frac{ 64sin55^{\circ}{10}'  }{9.8}

     = 5.36 m

    This is wrong. the answer says 5.29 m.

    2. The velocity of a point moving along the x axis is given by
     V^2 = 10 + 8x - 2x^2
    Show that it is simple harmonic.

     \ddot{x} = \frac{d}{dx}(\frac{1}{2}v^2)

     = \frac{d}{dx}(\frac{1}{2}10 + 8x - 2x^2)

     = \frac{d}{dx}5+4x - x^2

     = 4 - 2x

    However this is not the form -n^2x

    so how would I do this?

    Thanks!!! You guys are really helpful!
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  2. #2
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    Quote Originally Posted by differentiate View Post
    Could someone check these for me? i was told to put g = 9.8 m/s for all calculus questions, and I think it's a bit ridiculous.

    1. A long jumper running at 8 m/s jumps into the air, rising to a height of 0.7 metres. What is the length of the jump.

    maximum height
     0.7 = \frac{V^2sin^2\alpha }{2g}

     0.7 = \frac{8^2sin^2\alpha }{2\cdot 9.8}

     13.72 = 64sin^2\alpha

     sin\alpha = 0.463006479

    \alpha = 27^{\circ}{35}'

    Range
     \frac{V^2sin2\alpha }{9.8}

     =\frac{ 64sin55^{\circ}{10}'  }{9.8}

     = 5.36 m

    This is wrong. the answer says 5.29 m.


    2. The velocity of a point moving along the x axis is given by
     V^2 = 10 + 8x - 2x^2
    Show that it is simple harmonic.

     \ddot{x} = \frac{d}{dx}(\frac{1}{2}v^2)

     = \frac{d}{dx}(\frac{1}{2}10 + 8x - 2x^2)

     = \frac{d}{dx}5+4x - x^2

     = 4 - 2x

    However this is not the form -n^2x

    so how would I do this?

    Thanks!!! You guys are really helpful!
    (1) why so upset over a difference of 7 cm ? ... whoever came up with that "answer" used 10 for g in their calculations.

    (2) v^2 = 10+8x-2x^2

    2v \frac{dv}{dt} = (8-4x)\frac{dx}{dt}

    a = 4-2x = -(\sqrt{2})^2 \cdot (x-2)

    the equilibrium position is x = 2 instead of the "usual" x = 0

    the motion is simple harmonic
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  3. #3
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    hey thanks skeeter! thats awesome
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