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Math Help - Maximum displacement

  1. #1
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    Exclamation Maximum displacement

    On the particle displacement-time graph x=3sin(pie/8)t in units of centimetres and seconds. Find the first two times when the displacement is maximum

    Thanks for any help
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by nath_quam
    On the particle displacement-time graph x=3sin(pie/8)t in units of centimetres and seconds. Find the first two times when the displacement is maximum

    Thanks for any help
    for x to be maximum
    sin(pie/8)t = +1 or -1
    sin(pie/8)t = +1 = sin(pie/2)
    t/8= 1/2
    t=4sec.
    sin(pie/8)t =-1 =sin(3pie/2)
    t/8 =3/2
    t=12sec.
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  3. #3
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    Quote Originally Posted by nath_quam
    On the particle displacement-time graph x=3sin(pie/8)t in units of centimetres and seconds. Find the first two times when the displacement is maximum

    Thanks for any help
    Hello,

    1. Derivate x(t) with respect to t:

    {dx\over dt}={3\over 8}\cos\left({\pi\over 8}\cdot t \right)

    You'll get a maximum if dx/dt = 0 and the 2nd derivative is negative:

    {d^2x\over dt^2}={3\over 64}\left(-\sin\left({\pi\over 8}\cdot t \right) \right)

    dx/dt = 0 if \left({\pi\over 8}\cdot t \right)={{2k+1}\over 2}\cdot \pi

    That means t = 4*(2k+1)

    You'll get a maximum if k = 1, 3, 5, ... (Use the 2nd derivative to sort these numbers out!)

    So the first 2 maxima ar at t = 4 or t = 20. (See attachment)

    Greetings

    EB
    Attached Thumbnails Attached Thumbnails Maximum displacement-sin_max.gif  
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  4. #4
    Super Member malaygoel's Avatar
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    Quote Originally Posted by earboth
    Hello,

    1. Derivate x(t) with respect to t:

    {dx\over dt}={3\over 8}\cos\left({\pi\over 8}\cdot t \right)

    You'll get a maximum if dx/dt = 0 and the 2nd derivative is negative:

    {d^2x\over dt^2}={3\over 64}\left(-\sin\left({\pi\over 8}\cdot t \right) \right)

    dx/dt = 0 if \left({\pi\over 8}\cdot t \right)={{2k+1}\over 2}\cdot \pi

    That means t = 4*(2k+1)

    You'll get a maximum if k = 1, 3, 5, ... (Use the 2nd derivative to sort these numbers out!)

    So the first 2 maxima ar at t = 4 or t = 20. (See attachment)

    Greetings

    EB
    will t=12 also not correspond to maximum displacement?
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  5. #5
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    Quote Originally Posted by malaygoel
    will t=12 also not correspond to maximum displacement?
    Hello,

    you can calculate extreme values: the minimum, that's the smallest, and the maximum, that's the greatest.

    So in my opinion the value you get at t = 12 is a minimum and not a maximum.

    Greetings

    EB
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  6. #6
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    Thanks for your help guys but because the time axis is classed as the origin wouldn't the first two be 4 and 12 because its the absolute distance/maximum from the origin.
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  7. #7
    Forum Admin topsquark's Avatar
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    The solution probably depends on if you are doing Math or Physics. It sounds like a Physics problem, so I'd say t=4, 12. If it's a Math problem I'd say it's t=4, 20. (Physics is a lot more liberal in it's terms: maximum displacement implies largest distance from the origin in any direction, though that's not what the Mathematical definition of it would be.)

    -Dan
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