Question

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?

Answer 1

For the units of c, you have
\[N\cdot s/m\]
For the unist of k, you have,
\[N/m\]

Answer 2

sorry, the equation is

\[m \frac{ d^2y }{ dt^2 } +c \frac{ dy }{ dt } +ky = f(t)\]

Answer 3

John_ES, how do I get that?

Answer 4

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).

Answer 5

ohh, so the f(t) is in newton?

Answer 6

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.

Answer 7

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,
\[N=Kg\cdot\frac{m}{s^2}\]

Answer 8

aahh, I see... Thank you very mcuh John_ES. :)