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Math Help - Fast SHM question

  1. #1
    Junior Member Misa-Campo's Avatar
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    Fast SHM question



    thanks in advance
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi !
    Quote Originally Posted by Misa-Campo View Post
    Note that \sin(4t)+\sqrt{3}\cos(4t)=2\left( \frac{1}{2}\sin(4t)+\frac{\sqrt{3}}{2}\cos(4t)\rig  ht)<br />
  =2\left( \cos\frac{\pi}{3}\sin(4t)+\sin\frac{\pi}{3}\cos(4t  )\right). As \sin a \cos b+\sin b \cos a =\sin(a+b), we get that \sin(4t)+\sqrt{3}\cos(4t)=2\sin\left( 4t+\frac{\pi}{3}\right) hence x(t)=3+2\sin\left( 4t+\frac{\pi}{3}\right). Can you use this to answer the two questions?
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  3. #3
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    Quote Originally Posted by Misa-Campo View Post


    thanks in advance
    By re-writing as x = 3 + 2 \sin(4t + \phi) it's simple to see that the centre of motion is x = 3.

    i) Solve x = 3 for t: \tan (4t) = -\sqrt{3} etc.

    ii) v = dx/dt = 8 \cos (4t + \phi). Clearly the maximum value is v = 8.
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  4. #4
    Junior Member Misa-Campo's Avatar
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    Quote Originally Posted by mr fantastic View Post
    By re-writing as x = 3 + 2 \sin(4t + \phi) it's simple to see that the centre of motion is x = 3.

    i) Solve x = 3 for t: \tan (4t) = -\sqrt{3} etc.

    ii) v = dx/dt = 8 \cos (4t + \phi). Clearly the maximum value is v = 8.
    for i) im still stuck

    what do you mean by solve x = 3 for t: \tan (4t) = -\sqrt{3} etc. I've only started this topic so im not too sure
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  5. #5
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    Quote Originally Posted by Misa-Campo View Post


    thanks in advance
    Quote Originally Posted by mr fantastic View Post
    By re-writing as x = 3 + 2 \sin(4t + \phi) it's simple to see that the centre of motion is x = 3.

    i) Solve x = 3 for t: \tan (4t) = -\sqrt{3} etc.

    [snip]
    Quote Originally Posted by Misa-Campo View Post
    for i) im still stuck

    what do you mean by solve x = 3 for t: \tan (4t) = -\sqrt{3} etc.
    Since the centre of motion is x = 3, you want the times at which x = 3:

    3 = 3 + \sin (4t) + \sqrt{3} \cos (4t)

    \Rightarrow 0 = \sin (4t) + \sqrt{3} \cos (4t)

    \Rightarrow - \sin (4t) = \sqrt{3} \cos (4t)

    \Rightarrow - \tan (4t) = \sqrt{3}

    \Rightarrow \tan (4t) = - \sqrt{3}.

    And you know how to solve this trig equation ....?
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