Identical massless springs?

Question

particle is attached between two identical massless springs on a horizontal frictionless table. Both springs swivel freely at the fixed pins with no friction. Wach spring has a natural length of L, a force constant k, and is unstressed when the particle is at the origin.

Looks like: (The 1s are supposed to be the two springs)
.
1
1
@
1
1
.

If the particle is constrained to move only along the x-axis, what is potential energy? (we don't care about the constraint force that keeps y=0 at all times)

The x-direction is beside the @ (---x---)

Also, find the net force actoin on the particle for the case above.

Help with any part of this would be wonderful, I am so lost.

Answer 1

All you have to do is write an equation for the displacement of the springs as a function of x. When the the particle is pulled to any x, the springs form 2 right triangles with the x and y axes:

\
\
@
/
/

so lets call the length of the stretched springs h and define the displacement d = h-L:
h^2 = x^2 + y^2
y = L
h^2 = x^2 + L^2
h = sqrt(x^2 + L^2)
d+L = sqrt(x^2 + L^2)
d = sqrt(x^2 + L^2) - L

potential in a spring is equal to 1/2 times the constant times the displacement squared. So for one spring:

PE = 0.5*k*d^2
and for both springs, double it:
PE = k*d^2
PE = k*(sqrt(x^2 + L^2) - L)^2

To find the forces you have to do some more trig. The total force of one spring equals k times the displacement. In the y direction, you know that the net forces will always be 0 because the two springs cancel out. In the x direction, the x components will be added. x components are found be multiplying the magnitude times cos(theta) where theta is the angle formed. So:

F_y = 0
F_x = k*d*cos(theta) + k*d*cos(theta)
F_x = 2*k*d*cos(theta)
F_x = 2*k*(sqrt(x^2 + L^2) - L)*cos(theta)

cos(theta) = adjacent/hypotenuse
cos(theta) = x/h
cos(theta) = x/(d+L)
cos(theta) = x/(sqrt(x^2 + L^2) - L +L))
cos(theta) = x/sqrt(x^2 + L^2)

F_x = 2*k*(sqrt(x^2 + L^2) - L)*x/sqrt(x^2 + L^2)

Answer 2

As you move the particle 'left' and 'right' on the surface, what happens? The springs strech...they must since now the particle to what they are attatched is farther away from their piviot point than their normal length of L. Now the particle is at some distance x from the origin and is still some distance L away from the springs' piviot point. The resultant distance is the new, streched, length of the springs.

When the springs strech they exert some pulling force back on the object in the opposite direction. In this case, since there are two identicle, symetrical spings (one pulling "up" at an angle, another pulling "down" at an angle), the vertical components of their forces cancel leaving only a horizontal net force acting on the object.
Because the system is symetrical amount an imaginary Y axis running down the center, the total horizontal new force equals twice the horizontal force supplied by a single spring.

What you need you need to do to solve this problem is compare the springs stretch length to their normal length by fining the resultant distance between the spring's piviot point and the location of the particle.
R = sqrt (X^2 + L^2)
Now find out how far the spring has stretched by subtracting the normal length,
d = R - L
Where d is the displacement distance the string was streched.

Once you know d, you can find the force the spring exerts.
F = k*d
(the direction is along the spring, back towards its piviot point)

Now you know the total force per spring, but you need to know the horizontal component of the force, so you need to know the angle with which the force acts.
If you draw out the triangles you can see that you need to find the sine of the angle (theta) between the vertical and the resultant line to where the particle is located. The sine of theta equals the x displacement divided by the resultant.
Sin (theta) = X / R

So now combining everything together,
the horizontal component of the force exerted by one spring is:
F_x = k*(R - L) * (X / R) = k*X * (1 - L / R)
F_x = k*X * (1 - L / sqrt (X^2 + L^2))

And the total, net, force acting on the particle is twice this amount,
2*k*X * (1 - L / sqrt (X^2 + L^2))

Answer 3

First, the force: The angle each spring makes with the y axis is θ = arcsin x/L (sin¹ (x/L)). The extended length is L/cosθ. So, the chnage in length is L(1/cosθ - 1), and the spring tension T = kL(1/cosθ - 1)
The compnent of the tension in the x direction, Fx = Tsinθ. Putting in the previous value of T→ Fx = kL(tanθ - sinθ). Since there are 2 springs, the total Fx = 2kL(tanθ - sinθ). Since sinθ = x/L,
Fx = 2kL(x/Lcosθ - x/L) = 2kx(1/cosθ - 1)

The energy is defined as ∫Fx dx. I'm not up to doing the integration involving cos(sin¹(x/L)), but that's how it works out.

Good luck

Answer 4

it rather is pleasing, yet i've got not got time. The 5 in sequence have some thing like 5 instances the era oscillation of a unmarried spring, because of the fact the displacement distance is longer for the comparable tension. in case you start up up with the unmarried spring, then, and v/M some thing or different, then you are lowering the compression distance for the comparable tension to one million/5th of that by making use of paralleling 5 of them, so i might wager that is approximately 25 instances as quickly because of the fact the 1st one, yet i'm beneficial it rather is incorrect. have relaxing fidgeting with your springs and balls.