来自《耶鲁大学开放课程：基础物理 第16集 第16集》

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  You can pick some other crazy origin, but it doesn't help.
  你可以选其它诡异的原点 但会很麻烦
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  So, we'll assume φ is 0 in this problem.
  因此对这个问题我们假定 φ 为零
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  If that is x, what is velocity at time t?
  如果这是 x 那么 t 时刻的速度是多少
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  Take the derivative of this, you get -ω·A·sin(ω·t).
  对这个求导 得到 -ω·A·sin(ω·t)
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  That means the velocity is also oscillating sinusiodally,
  这意味着速度也在以正弦方式振荡
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  but the amplitude for oscillation is ω times A.
  但振幅是 ω 乘以 A
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  So, if A is the range of variation for x,
  如果 A 是 x 的变化范围
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  ω times A is the range of variation for velocity.
  ω 乘以 A 就是速度的变化范围
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  Velocity will go all the way from plus ω·A to -ω·A.
  速度会一直在正 ω·A 和 -ω·A 之间变化
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  The acceleration, which is one more derivative,
  加速度要再求一次导数
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  is -ω^2·A·cos(ω·t),
  是 -ω^2·A·cos(ω·t)
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  which is really -ω^2 times x itself.
  也就是 -ω^2 乘以 x