A mass attached to two horizontal springs, vertical motion

In summary: Please note: in your...In summary, the mass is connected to two springs, one with a constant force and the other with a force that depends on the angle the mass is at when it is displaced. The mass is in equilibrium when the two forces are equal. The equation of motion is found to be: F=-kx, a(x)=-(w^2)(x), and the frequency of oscillation is (w/m)x.
  • #1
roman15
70
0

Homework Statement


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.
| |
|______m______|
| |
this is what the system looks like, with each string 2L in the picture. The x direction is vertical


Homework Equations


F=-kx
a(x)=-(w^2)(x)

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)
 
Physics news on Phys.org
  • #2
roman15 said:

Homework Statement


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.
| |
|______m______|
| |
this is what the system looks like, with each string 2L in the picture. The x direction is vertical
I am having difficulty understanding the problem. Are the two springs in series or parallel? Are they both above the mass?

AM
 
  • #3
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
 

Attachments

  • physics pic.gif
    physics pic.gif
    2.1 KB · Views: 1,064
  • #4
roman15 said:
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
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
i don't 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 isn't a horizontal surface
 
  • #6
roman15 said:
i don't 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 isn't a horizontal surface
Ok. We will assume it is in orbit in space
from examining the forces I got that a(x)=(-2kxsin(theta))/m
i didnt know what to do next
would w=squareroot(2k/m)
Yes.

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

[tex]x = A\sin{(\omega t + \phi)}[/tex]

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

What is the value of A?

AM
 
  • #7
wait, wouldn't 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 I am not sure what A would be
 
  • #8
roman15 said:
wait, wouldn't 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)
No. If you take the second derivative of x with respect to time you get the acceleration:

[tex]\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[/tex]

You have found that:

[tex]a = -2kx/m[/tex]

So that means that:

[tex]\omega^2 = 2k/m[/tex]

and I am not sure what A would be
What is the maximum value of x?

AM
 
Last edited:
  • #9
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
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
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
roman15 said:
from examining the forces I got that [itex]a(x)=(-2kx\sin{(\theta))}/m[/itex]
Please note: in your original answer the sin(theta) term is incorrect. I am not sure what [itex]\theta[/itex] is. As I have stated above, the correct equation for the acceleration of m is:

a(x)=-2kx/m

where x = [itex]\Delta x[/itex] - the displacement from the equilibrium position in the vertical direction.

AM
 
  • #13
hmm I am 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
roman15 said:
hmm I am 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)
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
no the question doesn't 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
SammyS said:
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.
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: [tex]F_s=k\left(\sqrt{4L^2+x^2}\,-\,L\right)[/tex]

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.

[tex]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)[/tex]

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

.
 
  • #17
roman15 said:
no the question doesn't say that the springs are vertical, and from the diagram i gave shows that the springs are horizontal, but the x direction is vertical
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 [itex]\theta[/itex]. 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.:

[tex]F_x = -k(2L)sin\theta + -k(2L)sin\theta = -4kLsin\theta[/tex]

Since [itex]sin\theta = x/2L[/itex] the equation becomes:

[tex]F_x = -2kx[/tex]

So:

[tex]a = -2kx/m[/tex]

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

AM
 
Last edited:
  • #18
My derivation come out to F = -2kx(1/2+...) ≈ -kx.

Check my derivation. I'll look later.
 
  • #19
SammyS said:
My derivation come out to F = -2kx(1/2+...) ≈ -kx.

Check my derivation. I'll look later.
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
Andrew Mason said:
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 [itex]\theta[/itex]. 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.:

[tex]F_x = -k(2L)sin\theta + -k(2L)sin\theta = -4kLsin\theta[/tex]

Since [itex]sin\theta = x/2L[/itex] the equation becomes:

[tex]F_x = -2kx[/tex]

So:

[tex]a = -2kx/m[/tex]

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

AM

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
 
  • #21
are you assuming that the change in the length of the springs is equal to the change in x?
and if so then wouldn't it be squareroot(4L^2-x^2) - 2L
 
  • #22
oh wait sorry i didnt see that there was a second page to this message lol, ignore what i just said
ok this makes sense now, because today i went to my physics tutorial and my TA said that we had to show that force is this case was essential equal to -kx
thanks to the both of you for the help!
 
  • #23
SammyS said:
My derivation come out to F = -2kx(1/2+...) ≈ -kx.

Check my derivation. I'll look later.

i think is this the right way, cause i asked my professor and he mentioned making an approximation and having to ignore higher orders of x

just one question, why is the change in L = to the change in x, cause that's what you guys were doing right?
 
  • #24
roman15 said:
i think is this the right way, cause i asked my professor and he mentioned making an approximation and having to ignore higher orders of x

just one question, why is the change in L = to the change in x, cause that's what you guys were doing right?
No. The change is in the direction of L by a vertical displacement by x. The vertical force is the vertical component of the spring force, which is still mostly horizontal except for that small vertical component which is [itex]-kLsin\theta[/itex] for each spring:

[tex]F_x = -k(L)sin\theta + -k(L)sin\theta = -2kLsin\theta[/tex]

Since [itex]sin\theta = x/2L[/itex] the equation becomes:

[tex]F_x = -kx[/tex]


AM
 
  • #25
oh ok that makes sense then
thanks so much!
 
  • #26
im curious now, i haven't learned how to do power series expansions, would you mind explaining how to do it, i tried looking online but couldn't find a simple explanation for it
 
  • #27
roman15 said:
im curious now, i haven't learned how to do power series expansions, would you mind explaining how to do it, i tried looking online but couldn't find a simple explanation for it

Here are a couple of links:

http://mathworld.wolfram.com/SeriesExpansion.html" [Broken].

http://en.wikipedia.org/wiki/Power_series" [Broken]

Personally, I liked Andrew Mason's approach better than what I was doing with the full blown power series thing. That's one reason I quit responding once the configuration was settled.
 
Last edited by a moderator:

1. What is the basic concept behind a mass attached to two horizontal springs with vertical motion?

The basic concept is that a mass is attached to two horizontal springs and can move vertically due to the stretching and compressing of the springs.

2. How do the two horizontal springs affect the vertical motion of the mass?

The two horizontal springs work together to determine the frequency and amplitude of the vertical motion. They act as a restoring force, pulling the mass back towards the equilibrium position when it is displaced.

3. What is the equation of motion for a mass attached to two horizontal springs with vertical motion?

The equation of motion is m × a = -k1x - k2x, where m is the mass, a is the acceleration, k1 and k2 are the spring constants, and x is the displacement of the mass from the equilibrium position.

4. What factors can affect the vertical motion of the mass in this system?

The vertical motion of the mass can be affected by the mass of the object, the spring constants, the amplitude of the motion, and any external forces acting on the system.

5. How can this system be used in real-world applications?

This type of system can be used in many real-world applications, such as in shock absorbers for vehicles, in seismometers for measuring earthquakes, and in suspension systems for buildings or bridges.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
274
  • Introductory Physics Homework Help
Replies
31
Views
971
  • Introductory Physics Homework Help
Replies
2
Views
939
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
24
Views
891
  • Introductory Physics Homework Help
Replies
3
Views
830
  • Introductory Physics Homework Help
Replies
8
Views
4K
  • Introductory Physics Homework Help
Replies
27
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
2K
Back
Top