Double Spring Mass System - Maximizing Velocity

In summary, the conversation discusses a problem involving two springs and two masses connected in sequence, and the goal of finding the optimal time to release the right spring to maximize the peak velocity of the right block. It is suggested to solve the system of equations analytically using Runge-Kutta or energy conservation, and a possible solution is also provided using normal coordinates. However, it is also mentioned that the problem may not have a simple solution due to the dependence of displacements on each other.
  • #1
ejhong
2
0
Looking for some pointers on how to approach this problem:

I'm considering a system like the following - two springs and two masses connected in sequence:

http://www.myphysicslab.com/dbl_spring1.html

I would like to find the following:

Given m1, m2, k1, k2. The springs are compressed each by a certain amount (x1start and x2start) and the velocities are set to 0. The springs can be released at any point in time. Once a spring is released and it stretches so that its stretch reaches a peak, it is clamped at that position. Once both springs reach their peak stretch the simulation is ended. Here is what I'm trying to find:

Assuming the left spring is released at t=0. At what time should the right spring be released to maximize the peak velocity of the right block.

I could iterate numerically using Runge-Kutta and try stepping up the release time of the right spring and record peak velocity of the right block but that seems very brute forced. I would like to find a better solution for this if possible.

As a side question, I was wondering if there was an analytical solution to this system?

Thanks for any pointers.
 
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  • #2
As a side note: If you are looking at "optimized" velocity acceleration using a spring system, consider going "out-of-the-box"

I have worked on a similar problem, and concluded through my work that a "conical" spring works extremely well, especially if one attaches "tension stage" kevlar filaments in a certain way. I can not go into more detail, as such a design is nothing short of a powerful gun without using explosives.

Your on the right track, but allow yourself to be creative.
 
  • #3
You can solve the system that you linked to analytically. I would simplify the system by making x_1 and x_2 measured from the equilibrium points of the masses (0 for both when the whole system is in equilibrium.. both springs at their rest lengths). Then the equations are:
[tex]m_1 \ddot{x_1}+(k_1+k_2)x_2-k_2 x_2=0[/tex]
[tex]m_2 \ddot{x_2}+k_2 x_2-k_2 x_2=0[/tex]

To see how to go about an analytic solution you can look in Marion Thornton's Classical Dynamics of Particles and Systems chapter 12 (coupled oscillations) though the example they give is simpler because it has a lot of symmetry. The solution is basically the same, but your frequencies will be much messier.

Basically you guess a solution of the form:
[tex] x_1(t)=B_1 e^{i \omega t}[/tex]
[tex] x_2(t)=B_2 e^{i \omega t}[/tex]
plug them into the equations above and get a determinant of the coefficients of B_1 and B_2. You'll actually be getting four frequencies (two each of the same magnitude, but different signs) because of a square root. You end up with a final solution of the form:
[tex]x_1(t)=B_1^{+} e^{i \omega_1 t}+B_1^{-} e^{-i \omega_1 t} +B_2^{+} e^{i \omega_2 t} +B_2^{-} e^{-i \omega_2 t}[/tex]
[tex]x_2(t)=-B_1^{+} e^{i \omega_1 t}-B_1^{-} e^{-i \omega_1 t} +B_2^{+} e^{i \omega_2 t} +B_2^{-} e^{-i \omega_2 t}[/tex]
where the B coefficients are determined by the initial conditions and the frequencies are just the two magnitudes of the four total frequencies (if I said that correctly). Anyway, there you go. You can also find a set of normal coordinates, but I don't know if that will really help. Once m_1 is stopped, the other mass will just behave like a simple oscillator so it should be pretty simple from there.
 
  • #4
ejhong said:
As a side question, I was wondering if there was an analytical solution to this system?

Thanks for any pointers.


It look that after you reduced the system of linear second order DE to a system of linear first order DE, your system of equations can be written as a linear matrix equation

[tex]Y' = AY + F[/tex]

where
[tex]A=\left(\begin{array}{cccc} 0& 0& 1&0\\ 0&0&0&1\\
-\frac{(k_1 + k_2)}{m_1}& \frac{k_2}{m_1}&0&0\\
\frac{k_2}{m_2}& -\frac{k_2}{m_2}&0&0\end{array} \right) ,

Y=\left(\begin{array}{c} x_1\\ x_2\\ v_1\\ v_2\end{array}\right)
\mbox{ and }
F=\left(\begin{array}{c} 0\\ 0\\ \frac{k_1R_1}{m_1} - \frac{k_2(w_1+R_2)}{m_1}\\ \frac{k_2(w_1+R_2)}{m_2}\end{array}\right) [/tex]

if I copied correctly.

I think the matrix equation can be solve for Y by diagonalizing the matrix A. But haven't try yet.
 
  • #5
I would think that you could answer the question using energy conservation much easier than actually solving the differential equation. The maximum velocity will be achieved if, in the end, the two springs are uncompressed (and unstretched), and the other mass is stationary.
 
  • #6
DaleSpam said:
I would think that you could answer the question using energy conservation much easier than actually solving the differential equation. The maximum velocity will be achieved if, in the end, the two springs are uncompressed (and unstretched), and the other mass is stationary.

I'm not that sure how easy this problem can be solve using the conservation of mechanical energy. I thought from the conservation law you only get some relations involving velocities and displacements. The original problem post by ejhong has time factor.


The equation of motion is a system of coupled DE. I think it is only a hypothesis that the right block attain maximum speed when the two springs are uncompressed. How do we know that this situation is really a particular solution to the problem? The displacements of the two masses are not independent of each other.
 

1. What is a double spring mass system?

A double spring mass system is a mechanical system consisting of two springs and a mass that is attached to the springs. The mass is able to move vertically, stretching and compressing the springs. This system is commonly used in physics experiments to study oscillations and harmonic motion.

2. How can the velocity of a double spring mass system be maximized?

The velocity of a double spring mass system can be maximized by increasing the amplitude of the oscillations. This can be achieved by increasing the initial displacement of the mass or by increasing the tension in the springs. Additionally, reducing the damping forces in the system can also increase the maximum velocity.

3. What factors affect the maximum velocity of a double spring mass system?

The maximum velocity of a double spring mass system is affected by several factors, including the mass of the object, the stiffness of the springs, the initial displacement of the mass, and the damping forces present in the system. These factors can be manipulated to increase or decrease the maximum velocity of the system.

4. How does the total energy of a double spring mass system affect its maximum velocity?

The total energy of a double spring mass system is directly related to its maximum velocity. As the system oscillates, energy is constantly being exchanged between potential energy stored in the springs and kinetic energy of the mass. The maximum velocity occurs when all of the potential energy is converted into kinetic energy.

5. What are some real-world applications of double spring mass systems?

Double spring mass systems have various real-world applications, including shock absorbers in vehicles, pendulum clocks, spring-loaded doors, and even bungee jumping. They can also be used in earthquake engineering to study the response of structures to seismic forces. In the medical field, double spring mass systems are used in prosthetics and orthotics to mimic natural movements of the human body.

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