How Can the Spring Equation for a Chain of Springs be Solved?

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In summary, the conversation discusses a chain of springs and an equation of Newton's second law involving tension force. The conversation then explores finding the solution to an ODE using direct integration or Fourier transform, but the individual becomes stuck and asks for help. There is also a clarification about the equation involving k = 2 springs.
  • #1
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There is a chain of springs, such that qk is the displacement of the kth mass in this chain.

We have the next equation of Newton second law, where T is the tension force:
[tex]\frac{d^2q_k}{dt^2}=T(q_{k+1}-q_k)-T(q_k-q_{k-1})[/tex]
Let Qk=qk+1-qk

I showed that if Q(t)k=Q(k-ct)=Q(s)

then we get the next ODE:
(1)[tex] c^2 Q''(s)=T(s+1)-2T(s)+T(s-1)[/tex]
I need to show that the solution is:
[tex]Q(s)=\int_{s-1}^{s+1} (1-|s-z|)T(z)dz[/tex]
by direct integration i.e integrating (1) or by Fourier transform.

I tried both but I get stuck, by direct integration I get:

[tex]c^2Q(s)= \int^{s} \int^{x} [T(z+1)-2T(z)+T(z-1)] dz dx[/tex]
don't know how to proceed from here, can anyone give a helping hand? (-:
 
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  • #2
I also tried the Fourier transform I got to:
[tex]Q(s)=FT(FT^-1(T(s))*\frac{sin^2(w/2)}{(w/2)^2})[/tex]

I tried mathematica to solve this, but I am not sure how to write it with an implicit function as FT^-1(T(s)).
 
  • #3
I wonder about equation (1). Since:

[tex]\frac{d^2Q_k}{\text{dt}^2}=T\left(Q_{k+1}\right)-2T\left(Q_k\right)+T\left(Q_{k-1}\right)[/tex]

So, we have:

[tex]c^2 Q''(s)=T(Q(s+1))-2T(Q(s))+T(Q(s-1))[/tex]

instead of

[tex]c^2 Q''(s)=T(s+1)-2T(s)+T(s-1)[/tex]

Please clarify.
 
  • #4
What are the equations if you have k = 2 springs? Using your equations you would have displacements q0, q1, q2, and q3 involving 4 displacements and only two equations.
 
  • #5


I would say that the spring equation described here is a representation of the motion of a chain of springs, where qk represents the displacement of the kth mass in the chain. The equation is derived from Newton's second law, where T is the tension force acting on the springs. By setting Qk=qk+1-qk, we can simplify the equation and obtain an ODE (1) that represents the motion of the chain.

The goal is to show that the solution to this ODE is given by Q(s)=\int_{s-1}^{s+1} (1-|s-z|)T(z)dz, where s is a constant. This can be done through direct integration or by using Fourier transform. However, both methods seem to have led to a dead end.

In this case, it may be helpful to consider consulting with other experts in the field or conducting further research to find alternative methods for solving this ODE. It is also important to carefully check all calculations and assumptions made in the process to ensure accuracy and validity of the solution. Science is a collaborative and ever-evolving field, and seeking help and further exploration is a crucial part of the scientific process.
 

What is the "Spring equation"?

The "Spring equation" is a mathematical model that describes the behavior of a spring when a force is applied to it. It is also known as Hooke's law and is expressed as F = -kx, where F is the force applied, k is the spring constant, and x is the displacement of the spring.

What is the significance of the "Spring equation"?

The "Spring equation" is significant because it is a fundamental law of physics that explains the behavior of springs and other elastic materials. It is also used in various fields such as engineering, mechanics, and electromagnetics to understand and design systems that involve springs.

How is the "Spring equation" derived?

The "Spring equation" is derived from experiments and observations made by Robert Hooke in the 17th century. He found that the force required to stretch or compress a spring is directly proportional to the displacement of the spring. This relationship is known as Hooke's law and is the basis for the "Spring equation".

What are the limitations of the "Spring equation"?

The "Spring equation" has some limitations as it assumes that the spring is perfectly elastic and obeys Hooke's law at all times. In reality, there are factors such as material fatigue, temperature, and stress that can affect the behavior of a spring and cause it to deviate from the equation's predictions.

Can the "Spring equation" be applied to all types of springs?

The "Spring equation" can be applied to most springs, as long as they behave elastically and follow Hooke's law. However, it may not accurately describe the behavior of complex or non-linear springs, such as those used in shock absorbers or torsion springs. In these cases, more advanced equations may be needed to accurately model their behavior.

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