# A mass attached to two horizontal springs, vertical motion

1. Mar 6, 2011

### roman15

1. The problem statement, all variables and given/known data
A mass is connected between two identical springs with the same spring constant, k. The equilibrium length of each spring is L, but they are stretched to 2L when the mass is in equilibrium. Ignoring gravity, find the equations of motion and then angular frequency of oscillation, when the mass is displaced by a small distance +/-x.
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|______m______|
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this is what the system looks like, with each string 2L in the picture. The x direction is vertical

2. Relevant equations
F=-kx
a(x)=-(w^2)(x)

3. The attempt at a solution
from examining the forces I got that a(x)=(-2kxsin(theta))/m
i didnt know what to do next
would w=squareroot(2k/m)

2. Mar 7, 2011

### Andrew Mason

I am having difficulty understanding the problem. Are the two springs in series or parallel? Are they both above the mass?

AM

3. Mar 7, 2011

### roman15

here is an attached picture of what the system looks like, sorry for the confusion
the only difference from this picture is that the mass moves up and down, and the vertical direction is the x axis

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4. Mar 7, 2011

### Andrew Mason

I am still puzzled. I don't see how you can "ignore gravity".

Do a freebody diagram for m in the supposed equilibrium position. What are the downward forces on m? What is the upward force? How can this be equilibrium?

If m is on a horizontal surface, you can stretch the springs by an equal amount and achieve equilibrium. But when you turn it vertically, the springs will not be stretched equally.

AM

5. Mar 7, 2011

### roman15

i dont know, but were asked to just ignore gravity so the only forces acting on it when you displace the mass are the two forces of the springs, and no there isnt a horizontal surface

6. Mar 8, 2011

### Andrew Mason

Ok. We will assume it is in orbit in space
Yes.

Your equation is a differential equation that has a general solution of:

$$x = A\sin{(\omega t + \phi)}$$

where $$\omega = \sqrt{2k/m}$$ and $\phi$ is a phase factor that depends on your choice of when t = 0.

What is the value of A?

AM

7. Mar 8, 2011

### roman15

wait, wouldnt w= squareroot(2ksin(theta)/m)?
because the sin(theta) i got for the acceleration just depends on the initial angle the mass starts from when displaced, then x = Asin(omegat + phi)
and im not sure what A would be

8. Mar 8, 2011

### Andrew Mason

No. If you take the second derivative of x with respect to time you get the acceleration:

$$\frac{d^2x}{dt^2} = A\frac{d}{dt}(\omega(\cos{\omega t + \phi})) = A\omega^2(-\sin{(\omega t + \phi})) = - \omega^2 x = a$$

You have found that:

$$a = -2kx/m$$

So that means that:

$$\omega^2 = 2k/m$$

What is the maximum value of x?

AM

Last edited: Mar 8, 2011
9. Mar 8, 2011

### SammyS

Staff Emeritus
If the motion is perpendicular to the springs, the solution is quite a bit different.

In your figure this would be the case if the mass moved in a direction into and then out of the page -- away from the viewer then toward the viewer.

10. Mar 8, 2011

### roman15

oh so the equation for the acceleration that i got is for if the spring moves forwards and backwards, not up and down?
what do i do to make it for the mass going up and down?

11. Mar 8, 2011

### SammyS

Staff Emeritus
If the mass moves a distance x from it's equilibrium position, then each spring stretches from 2L to √(4L2+x2).

The component of force in the x-direction is just x/√(4L2+x2) times the magnitude of the force.

12. Mar 9, 2011

### Andrew Mason

Please note: in your original answer the sin(theta) term is incorrect. I am not sure what $\theta$ is. As I have stated above, the correct equation for the acceleration of m is:

a(x)=-2kx/m

where x = $\Delta x$ - the displacement from the equilibrium position in the vertical direction.

AM

13. Mar 9, 2011

### roman15

hmm im going to ask my professor if we should assume the springs stay the same length when we displace the mass, because the question says were displacing it by small distances
and for the sin(theta), well if you displace the mass then the x component of the spring force would include sin(theta)

14. Mar 9, 2011

### Andrew Mason

Your question says that the displacement is in the vertical direction. If the springs are vertical, that means the springs compress and stretch vertically along the axes of the springs.

AM

15. Mar 9, 2011

### roman15

no the question doesnt say that the springs are vertical, and from the diagram i gave shows that the springs are horizontal, but the x direction is vertical

16. Mar 9, 2011

### SammyS

Staff Emeritus
The above quote goes with my previous post in which the mass moves perpendicular to the line along which the springs lie when the system is in equilibrium.

When the mass is displaced a distance x from equilibrium the magnitude of the force exerted by each spring is: $$F_s=k\left(\sqrt{4L^2+x^2}\,-\,L\right)$$

Take the component of force in the x-direction & multiply by 2 to get the magnitude of the net Force. The net force, F, is in the negative x-direction.

$$F=2k\left(\sqrt{4L^2+x^2}\,-\,L\right)\frac{x}{\sqrt{4L^2+x^2}}=2kx\left(1-\frac{L}{\sqrt{4L^2+x^2}}\right)=2kx\left(1-\frac{1}{\sqrt{4+(x/L)^2}}\right)$$

The power series expansion for $$\frac{1}{\sqrt{4+u^2}}$$ is $$\frac{1}{\sqrt{4+u^2}\,}= {{1}\over{2}}-{{x^2}\over{16}}+{{3 x^4}\over{256}}-{{5 x^6}\over{2048}}+\dots$$

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17. Mar 9, 2011

### Andrew Mason

A simple statement saying the springs are stretched horizontally and the mass is displaced vertically would have made it clear. But I see you put that in the title. Missed that. In any event, this is not a problem because the solution is the same:

The restoring force on m is the vertical component of the spring force when the mass is displaced a distance x. The spring force is constant in magnitude (F = -k2L) but changes in direction. It is only the vertical component that you are concerned about. Let the angle that the springs make with the horizontal be $\theta$. Since the springs are identical and have the same stretch and length, they provide the same force and make the same angle for a given displacement.:

$$F_x = -k(2L)sin\theta + -k(2L)sin\theta = -4kLsin\theta$$

Since $sin\theta = x/2L$ the equation becomes:

$$F_x = -2kx$$

So:

$$a = -2kx/m$$

and, therefore the solution is the one I gave you above.

AM

Last edited: Mar 9, 2011
18. Mar 10, 2011

### SammyS

Staff Emeritus
My derivation come out to F = -2kx(1/2+...) ≈ -kx.

Check my derivation. I'll look later.

19. Mar 10, 2011

### Andrew Mason

You do not have to take into account the additional length of the spring since we are assuming very small x (ie. x<<L). I was using 2L as the stretch of the springs but perhaps it should be L. I was not clear on that. If it is L then just divide by 2: a = -kx/m

AM

20. Mar 10, 2011

### SammyS

Staff Emeritus
Each spring is stretched to 2L, but the its equilibrium length is L so the magnitude of the force exerted by each spring is: |F| = k(s ‒ s0) = k(2L ‒ L) = kL.

From which both AM & I agree that a = ‒kx/m .

→ (d2x/dt2) = ‒kx/m