How Does Spring Force Affect Particle Movement on a Frictionless Surface?

[FOBoi1122]

A particle of mass m is attatched to 2 identical springs on a horizontal frictionless table. both springs have spring constant k and an unstretched length L the particle is pulled a distance x along a direction perpindicular to the initial configuration of the springs, show that the force exerted on the particle due to the springs is

F= -2kx(1-(L/(x^2+L^2)^.5))i


please help us, we've tried to use f=-kxcos(theta), where cos(theta) = x/(x^2+L^2) but we are unable to determine why x = (x^2+L^2)^.5 - L which is needed to solve the problem

 

[HallsofIvy]

PLEASE do not post the same question under different topics!

Here is the answer I posted under "General Physics":
The problem itself is not clear. I could interpret this as two springs in series with the "particle" attached at the end or with the two springs both attached to a wall and the particle between them.

Since in the first case it would be simpler to treat the two springs as a single spring, I suspect the second is intended.

It also is not stated clearly, but I will assume it, that the two springs are attached at points 2L apart so that the springs start in equilibrium position with the particle between them.

Now, as the particle is moved perpendicular to the wall, a distance x, we see two congruent right triangles. One leg is of Length L (from point of attachment to center) and the other is of length x so the hypotenuse has length √(L²+ x²). The hypotenuse is, of course, the full stretched length of one spring so the amount of stretch is √(L²+ x²)-L.

The total force due to one spring is k√(L²+ x²)-L but this is directed toward the point of attachment. If we break that into components, parallel and perpendicular to the wall, we see that the parallel forces of the two springs cancel each other while the ones perpendicular to the wall add.

You are correct that the pependicular force for each spring is F (x/√(L²+ x²)so the total force is twice that: (2k√(L²+ x²)-L)(x/√(L²+ x²).

The answer you give is that, with -i since it is directed back toward the wall.

 

[M98Ranger]

First of all, great job on attempting to use the formula for force due to a spring, F=-kx. However, in this case, the particle is being pulled in a direction perpendicular to the initial configuration of the springs, so we cannot simply use the displacement x as the distance from the equilibrium position.

To solve this problem, we need to consider the geometry of the situation. Let's draw a diagram to visualize it.

We have a particle of mass m attached to two identical springs with spring constant k and unstretched length L. The particle is pulled a distance x along a direction perpendicular to the initial configuration of the springs. Let's label the angle between the displacement vector and the horizontal axis as θ.

Now, let's consider the forces acting on the particle. The two springs will exert equal and opposite forces on the particle, each with a magnitude of F=-kx. However, these forces are not acting in the same direction. The force from the left spring is acting in the direction of the displacement, while the force from the right spring is acting in the opposite direction.

Using trigonometry, we can see that the component of the force from the left spring in the direction of the displacement is F_left = -kx cosθ, and the component of the force from the right spring in the direction of the displacement is F_right = kx cosθ.

Since the two forces are acting in opposite directions, we can simply subtract them to get the net force on the particle in the direction of the displacement:

F = F_left - F_right = -kx cosθ - kx cosθ = -2kx cosθ

Now, we need to find the value of cosθ. From the diagram, we can see that cosθ = x/(x^2+L^2)^0.5. Substituting this into the equation for F, we get:

F = -2kx(x/(x^2+L^2)^0.5)

Next, we need to simplify this expression. We can use the identity (a^2+b^2)^0.5 = c to rewrite (x^2+L^2)^0.5 as (x^2+(L^2)^0.5)^0.5. This allows us to write the expression as:

F = -2kx(1-(L/(x^2+L^2)^0.5))