# Function of equilibrium position in a system with two springs and a mass

Consider a mass $$m$$ with a spring on either end, each attached to a wall. The mass can freely move horizontally, without friction forces. Let $$k_1$$ and $$k_2$$ be the spring constants of the springs respectively with equilibrium position $$x_1$$ and $$x_2$$. Write a function for the equilibrium position $$x_0$$ of the system using $$k_1$$, $$k_2$$, $$x_1$$, $$x_2$$ and $$m$$.

Why can't we set the equilibrium position at $$x_0=0$$ since we are free to choose where the origin is?

• Did the question come with a diagram showing the coordinate system they want you to use? It looks like $x_0$ needs to be written relative to $x_1$ and $x_2$. Feb 4, 2022 at 11:09
• Also, I edited the post to focus on the concept you are confused about. The previous version made it sound more like you wanted someone to solve the exercise, which is not allowed on PSE Feb 4, 2022 at 11:10
• No, it doesn't. The problem is that I don't understand how to write the equilibrium point. Do you have any advice? Feb 4, 2022 at 11:19
• Then in the original problem you have to study the function, calculate the velocity... I know how to do that. But I can't write the equilibrium position. Feb 4, 2022 at 11:23

You do not need to worry about walls, the separation of the walls, the initial lengths of springs, etc.

Assume that the two springs and the mass lie along a horizontal line.

Let the free (right) end of the left unstretched spring be at position $$x_1$$ and the free (left) end of right unstretched spring be at position $$x_2$$ both relative to some origin on the horizontal line. For example, the origin could be at the fixed (left) end of the left hand spring where it is attached to the wall.

The equilibrium position of the mass when it is attached to both springs will be at $$x_0$$ which is somewhere between position $$x_1$$ and position $$x_2$$.
You now can find the extensions of the springs, the forces they exert on the mass, . . . . . .

Equilibrium position is where the body will (eventually) rest. Sure, you can put the origin of your coordinate system at the equilibrium position, but then the question is - what are coordinates of the left and right walls in this new coordinate system? In other words, you do not know where exactly to put this new origin.

Here are hints to solve for equilibrium position:

• place the origin where the left wall is
• let $$x$$ denote position of mass
• let $$L$$ denote distance between the two walls
• let $$L_1$$ and $$L_2$$ denote lengths of the two springs in relaxed state
• write expressions for spring forces in terms of $$x$$, $$L_1$$, $$L_2$$ and $$L$$

You must be careful with the signs of the spring forces - spring exerts force always in the direction of its relaxed position.

In equilibrium, vector sum of the two (spring) forces equals zero.

• They are $-2x_1$ and $2x_2$ (assuming that the first spring is on the left). Correct? But how it will help to write a function for the equilibrium point? Or it is wrong because we have to consider the lenght of the mass, too (that is not considerated in the problem)? Feb 4, 2022 at 11:36
• @user149240 I am not sure what exactly you mean by $-2 x_1$ and $2 x_2$, but you should express the equilibrium position in terms of mass position, which means you have only one variable $x$. Feb 4, 2022 at 11:47
• Sorry, but what are $L$, $L_1$ and $L_2$? And how is it even possible that the forces are opposite? For example, if we compress the first spring, it will give a force to the right, but the second spring will give a force to the right too. So the two forces never are in equilibrium, they are only if both of them are zero. Or am I wrong? Feb 4, 2022 at 11:49
• Please do not post solutions to exercises Feb 4, 2022 at 11:52
• As I have written just now, how is it possible that the sum of the two forces is zero, if they have the same direction? Feb 4, 2022 at 11:54 Instead of the wall at the right I put a mass $$~M~$$.

you have now two equations

forces on mass $$~m~$$

$$F_m=k_1\,(x_1-L_1)+k_2\,(x_1-x_2-L_2)=0$$

and force on mass $$~M~$$

$$F_M=k_2\,(x_2-x_1-L_2)=0$$

where $$~L_i~$$ are the springs length (spring pre load is $$~k_i\,L_i~$$)

from here you can obtain the equilibrium positions $$~x_{10}~,x_{20}~$$ both are not depending on the masses .