0
$\begingroup$

enter image description here

The equations of motion of the spring mass system with, m = 1

$ \ddot{y_1} = -k_1y_1 + k_1(y_2-y_1)$

$ \ddot{y_2} = -k(y_2-y_1) - ky_2$

My question is with the second term in the first equation. I don't understand why the second term is

$ y_2 - y_1$

and why the second term is added instead of subtracted.

I know the spring force is in the opposite direction of displacement, and only displacement of the spring is taken into account. But, since the position is defined as zero at equilibrium position of each mass, I'm confused at the subtraction. Shouldn't they be added together since, stretching the system in the positive direction would stretch the string in the $y_1$ and $y_2$

$\endgroup$

migrated from math.stackexchange.com Jan 17 '16 at 10:25

This question came from our site for people studying math at any level and professionals in related fields.

3
$\begingroup$

The force in the middle spring is $k(y_2-y_1)$ because it is lengthened when $y_1\lt 0$ or $y_2 \gt 0$. A positive sign on that force indicates that $m_1$ is pulled down and $m_2$ is pulled up. When $m_2$ is pulled down, $y_2 \gt 0$ and there is a downward force on $m_1$, so it will cause an increase in $\ddot {y_1}$. It will also cause an upward force on $m_2$ for the same reason.

$\endgroup$
  • $\begingroup$ This makes sense if I think about perturbing the system by pulling on m2 and then analyzing the forces that result on each mass, but is there an easier way to visualize which direction the forces will be in? I think that's the hardest part is figuring the sign. For instance, if I imagined pulling just the m1 in the positive y1 direction, I would get a compressive force from the second spring (which is positive). How do I analyze the system so I always end up with the right direction for the forces? $\endgroup$ – justthom8 Jan 17 '16 at 19:21
  • $\begingroup$ I don't know any way other than being careful. You look at the direction the force is applied and make sure the sign of the acceleration is correct. When you pull $m_1$ down, the force on it is up, so the acceleration is negative, and so on. $\endgroup$ – Ross Millikan Jan 17 '16 at 21:44
  • $\begingroup$ The problem for me is, if you were to take away the springs and analyze just one mass at a time, there is no way of knowing whether the spring will be in tension or compression at a given time. For the first mass, I can see pulling it in the positive y1 direction would give a force in the opposite direction from spring 1, but the second spring is up in the air then as to whether it is in compression or tension. Thanks for the explanation though. $\endgroup$ – justthom8 Jan 17 '16 at 22:28
  • $\begingroup$ You are correct that you don't know from the start whether the middle spring is in tension or compression, but you should be able to see that its force is $k(y_2-y_1)$, then reason that if $y_2-y_1$ is positive the spring is in tension. That will get you the sign for both masses. If the middle spring is in tension, which means $y_2-y_1 \gt 0$ it produces a downward force on $m_1$, so adds to $\ddot y_1$ and produces an upward force on $m_2$, so subtracts from $\ddot y_2$ $\endgroup$ – Ross Millikan Jan 17 '16 at 23:35
  • $\begingroup$ What makes the problem hard is we are using the wrong variables. If we define $x_1=y_1-y_2, x_2=y_1+y_2$ the equations decouple. The $x_1$ equation uses all three springs and the $x_2$ equation ignores the middle spring. These are the two modes of the system. $\endgroup$ – Ross Millikan Jan 17 '16 at 23:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.