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A spring-mass system has a spring constant of $\displaystyle\frac{3N}{m}$. A mass of $2$ kg is attached to the spring, and the motion takes place in a viscous fluid that offers a resistance numerically equal to the magnitude of the instantaneous velocity. If the system is driven by an external force of $(3\cos(t)-2\sin(3t))$ N, determine the steady-state response.


My study group came up with the following. Is this reasonable?

$$\begin{align} &k= \displaystyle\frac{3N}{m}\\ &m=\displaystyle\frac{2k}{g}\\ &2y''+ry'+3y=3\cos(3t)-2\sin(3t) \end{align}$$

How do I find $r$?

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  • $\begingroup$ $r$ measures the resistance, right? which is "numerically equal to the magnitude of the instantaneous velocity", right? and the instantaneous velocity is $y'$, right? $\endgroup$ – Gerry Myerson Dec 11 '11 at 23:22
  • $\begingroup$ It would be better to write $k= \displaystyle 3\frac{N}{m}$ and definitely to write $m=2\ kg$ as the first makes the units clear and the second get the k and g in the same place. $\endgroup$ – Ross Millikan Dec 12 '11 at 1:43
  • $\begingroup$ By the way, I think the symbol $m$ is being used for two different things here. I think $N/m$ stands for newtons per meter, while $m=2{\rm\ kg}$ (and not $m=2k/g$) means mass is two kilograms. But perhaps I misunderstand. $\endgroup$ – Gerry Myerson Dec 12 '11 at 11:56
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"a resistance numerically equal to the magnitude of the instantaneous velocity"

Isn't the text saying that $r=1$ ?

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$ry'$ is the resistance, which the problem states is equal to the magnitude of the instantaneous velocity. you should understand the differential equation: the $y''$ term is $F=ma$ a la newton. the $y'$ term is some force proportional to the velocity, like a resistance, the $y$ term is a force proportional to the displacement, i.e. the spring, and the non-homogeneous term is a driving force. anyway, i believe the problem is saying that $r=1$

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  • $\begingroup$ How do we draw the conclusion that 1 is equal to r? We don't have a function for the velocity. $\endgroup$ – user17366 Dec 11 '11 at 23:26
  • $\begingroup$ Yes, we do: $y$ is displacement, so (as I suggested in my original comment) $y'$ is velocity. $\endgroup$ – Gerry Myerson Dec 12 '11 at 11:52

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