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butterflydreamer

  • one year ago

SIMPLE HARMONIC MOTION QUESTION - Please help me LOL. I completely forgot how to solve these types of questions :( A ship needs 10m of water to pass down a channel safely. At low tide, the channel is 9m deep and at high tide it is 12m deep. Low tide is at 10 a.m and high tide at 5pm. Assume that the tidal motion is simple harmonic. period= 14hrs and amplitude = 1.5m **At what time can the ship safely proceed before midnight?

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  1. anonymous
    • one year ago
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    :)

  2. anonymous
    • one year ago
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    For the first two qs show that the 2nd derivative = -w^2 x differentiate twice n then manipulate to get into that form

  3. butterflydreamer
    • one year ago
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    Would this be the equation to use? \[x = 1.5 \cos (\frac{ \pi }{ 7 } + \alpha) + 10.5\] Then let t = 0 , x = 9 so... \[9 = 10.5 + 1.5\cos \alpha\] \[-1.5 = 1.5 \cos \alpha\] \[\cos \alpha = -1\] \[\alpha = \pi\] let x = 10 (because.. the ship needs 10m of water?) \[10 = 1.5 \cos (\pi t /7 + \pi) + 10.5\] \[-0.5 = 1.5 \cos (\pi t/7 + \pi) \] \[-\frac{ 1 }{ 3} = \cos (\pi t/7 + \pi) \] \[\cos ^{-1} (-1/3) = \frac{ \pi t }{ 7 } + \pi \]

  4. butterflydreamer
    • one year ago
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    \[\cos ^{-1}(\frac{ 1 }{ 3 }) + \pi = \frac{ \pi t }{ 7 } + \pi\] \[\frac{ \pi t }{ 7 } = \cos ^{-1} (\frac{ 1 }{ 3 }) \] \[t = \frac{ 7\cos ^{-1}(\frac{ 1 }{ 3 }) }{ \pi }\]

  5. butterflydreamer
    • one year ago
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    right? xD going out on a limb right now.

  6. butterflydreamer
    • one year ago
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    so my final answer is 12:45 pm and 9:15 pm but it doesn't feel correct :/

  7. butterflydreamer
    • one year ago
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    *** 12:44pm and 9:15 am

  8. dan815
    • one year ago
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    |dw:1436575570798:dw|

  9. dan815
    • one year ago
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    |dw:1436576262945:dw|

  10. butterflydreamer
    • one year ago
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    thanksss ^_^

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