anonymous
  • anonymous
Deterimne the units on c, k, and f(t) in md^2/dt^2 + cdy/dt + ky = f(t) if m is in kg, y is in meters, and t is in seconds. What should I do?
Physics
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SOLVED
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chestercat
  • chestercat
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John_ES
  • John_ES
For the units of c, you have \[N\cdot s/m\] For the unist of k, you have, \[N/m\]
anonymous
  • anonymous
sorry, the equation is \[m \frac{ d^2y }{ dt^2 } +c \frac{ dy }{ dt } +ky = f(t)\]
anonymous
  • anonymous
John_ES, how do I get that?

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John_ES
  • John_ES
You need to know that the first term represents a force, \[F=ma=m\frac{d^2y}{dt^2}\]whose unist are Newtons. Then, all terms must have the same units, because this is the only way the eqution would take sense. So, the second term must be someting like, \[c\cdot velocity=N\Rightarrow c\cdot m/s=N\Rightarrow c=N\cdot s/m\] The same for the third term, \[k\cdot distance=N\Rightarrow k\cdot m=N\Rightarrow k=N/m\]You wil recognize the last term because is related to an elastic force (Hooke's Law).
anonymous
  • anonymous
ohh, so the f(t) is in newton?
John_ES
  • John_ES
Yes, the eqution represents a driven oscillator (check the web for a brief review http://hyperphysics.phy-astr.gsu.edu/hbase/oscdr.html), and f(t) could be considered as the external force on the system.
John_ES
  • John_ES
You can also deduce that the first terms must be in Newtons because, \[m\cdot\frac{d^2y}{dt^2}\rightarrow [M]\frac{[L]}{[T]^2}\]Just the units enclosed in the Newton symbol, \[N=Kg\cdot\frac{m}{s^2}\]
anonymous
  • anonymous
aahh, I see... Thank you very mcuh John_ES. :)

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