Forced vibration on mass between two pull springs

[Kurt Couffez]

Consider a mass m with a prestressed pull spring on either end, each attached to a wall. Let k1 and k2 be the pull spring constants of the springs. A displacement of the mass by a distance x results in the first spring k1 lengthening by a distance x(and pulling in the - direction), while the second spring k2 is abbreviating by a distance x (and pulls in the positive direction).

The equation of motion then becomes:

Three questions:

1. Is it correct that

2. If the answer on question (1) is yes, what is ω if k2>k1

3. Is is possible to resonate with a force F=A.cos(ωt) en what would be the amplitude of this force.

 

[Khashishi]

What is the resting equilibrium position for each spring individually?
The way you have it written is at x=0, but you say something about prestressed in your description so it isn't clear.

 

[Kurt Couffez]

Lets say the resting equilibrium position for the springs are x₁ and x₂. The reason why they have to be prestressed is because they are pull springs. They can only have a force in one direction. They both are pulling at any time at the mass. When they are not prestressed then, at the position x=0 (with the springs attached) they would be pressed together.

 

[Kurt Couffez]

Lets say the resting equilibrium position for the springs are x₁ and x₂. The reason why they have to be prestressed is because they are pull springs. They can only have a force in one direction. They both are pulling at any time at the mass. When they are not prestressed then, at the position x=0 (with the springs attached) they would be pressed together.
View attachment 107716

I do not think it even matters. The forces of pre-stretched springs eliminate each other at all times.

 

[Khashishi]

It should be
$ma = (k_1+k_2)x$
not
$ma = (k_1-k_2)x$

You can show it using the Euler-Lagrange equation.
$L = \frac{1}{2}m v^2 - \frac{1}{2}k_2 (x_2-x)^2 - \frac{1}{2} k_1 (x-x_1)^2$
$\frac{\partial L}{\partial x} = k_2 (x_2-x) - k_1 (x-x_1) = x (k_1 + k_2) + k_1 x_1 + k_2 x_2$
If x=0 is the equilibrium position, then $k_1 x_1 + k_2 x_2 = 0$

 

[Kurt Couffez]

Why is in this the enegery of both springs negative? Spring 1 is pulling upwards, spring 2 is pulling downwards.

 

[Kurt Couffez]

Ok, I have discovered the error. The restoring force F₂ should also be upwards. Spring 2 also wants to bring the mass to its equilibrium position even if it is a pull spring.
The question around the negative energy of the spring is also solved.

Thanks.