# 2 spring 1 mass system, find the equation of motion I have such a scenario which I was given to find out the equation of motion of the mass. I drew the free body diagram of the mass, and realized the force towards the left side of the mass is essentially the given F in the picture and the left spring is almost nonexistent. Take $x$ to the left as positive. The free body diagram of the left spring is as such. where F is the same on both side according to hooke's law.

That is how the left side of the mass experience force F and the right side experience $-kx$ where $x$ is the distance from the static right dot (origin) to the mass. So, the equation of motion of the mass I found is:

$F-kx=ma$

However, I was given the equation otherwise:

$F-kx-kx=ma$

Tell me if I am wrong.

I learned that when 2 springs is connected in series, they experience the same force, and the effective k become: $\frac{k_1k_2}{k_1+k_2}$ and their $x$ added up to become: $F=\frac{k_1k_2}{k_1+k_2}(x_1+x_2)$ where $F=k_1x_1=k_2x_2$

So, when the springs are the same, the equation simplifies to $F=kx$ where 2 springs becomes 1 spring. I think this is what is happening here.

Am I correct? Or is the equation given to me is correct?

• Is the force being continuously applied? Mar 2, 2017 at 7:14
• the force is a sinusoidal function of time Mar 2, 2017 at 7:26
• If the force on the left spring is known to be $F$, then only the right spring matters. You can check a few limiting cases: $F=0$, or $k_2=0$, or $k_1\rightarrow\infty$. Mar 2, 2017 at 8:11
• I don't get the use of the limiting case since the 2 springs are identical. Mar 2, 2017 at 8:13
• @newbie, just treat the spring constants to be independent and see what you get. Mar 2, 2017 at 8:19

Update due to my not looking at the diagram carefully enough and assuming that both springs had one end fixed.

If a force is applied to the left as shown in the diagram the displacement/velocity/acceleration of the mass is only determined by the value of the force the mass and the spring constant of the right hand mass.

Springs connected to two fixed ends and the mass displaced.

The following assumes that the two ends of the springs with the same spring constant not connected to the mass are fixed and the mass is moved to one side.

It is easier, although no essential, to think of the two springs to be unstretched when the mass is in its static equilibrium position and allowing the springs to be compressed as well as being stretched.

In your diagram if the mass is moved to the right the right hand spring becomes compressed and so exerts a force $F$ on the mass to the left.
The left hand spring is stretched to the right and so exerts a force $F$ on the mass to the left.

So the net force on the mass due to the two springs is $2F$ to the left.

A similar analysis can be done if both springs are stretched when the mass is in the equilibrium position.

Moving the mass to the right means the right hand spring pulls less to the right (equivalent to more pull to the left) and the left hand spring pulls more to the left.

• For your info, the springs are initially not stretched. Then, they are stretched to the left by F from the origin and F is a function of time. Mar 2, 2017 at 8:08
• @newbie Your springs in series formula assumes that the springs are connected together with the mass at one of the ends. So both springs stretch by the same amount. This is not the case in this problem where you have one spring stretched and the other compressed. Mar 2, 2017 at 8:12
• Shouldn't they stretched by an equal amount in this case since the 2 springs are identical? How is one got stretched and another got compressed? Mar 2, 2017 at 8:15
Your free body diagram of left spring is wrong, the spring experiences force of $F+kx$ on both sides. The left spring will also be stretched by distance x.