How Do You Determine the Frequency of a Mass Connected to Four Springs?

In summary, the problem involves a mass connected to 4 springs forming a square with sides √2 a. The springs have spring constants k and natural length a/2. When the mass is displaced by d << a towards one of the corners, the frequency is given by √(3k/m). By setting up force equations and using similar triangles, the vertical force on each of the springs is determined to be kd/2, leading to the solution of ma = -3xk.
  • #1
Aaron7
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Homework Statement


The mass is connected to 4 springs, each connected to a corner of a square with sides √2 a.
The springs have spring constants k and natural length a/2.

Show the frequency of the mass when it is displaced by d << a towards one of the corners is √(3k/m)

Homework Equations


F=kx

The Attempt at a Solution



When in equilibrium the springs are length a.
The forces acting in the same direction of motion would be (a/2 - d)k and -(a/2 + d)k since the spring is already stretch by a/2.

I am having trouble finding the vertical forces acting on the mass by the other 2 springs.
The length of the spring would be stretched to √(a^2 + d^2) so the force would be (√(a^2 + d^2) - a/2)k but I need to find the vertical component of this force.

Working backwards I find that the vertical force on each of the springs must be kd/2 so that (a/2 -d)k - (a/2 +d)k - kd/2 - kd/2 => -3dk so I can move on to use ma = -3xk etc to find the answer. I see a link between being stretched twice its length already and having k/2 but I am not sure why.

Many thanks.
 
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  • #2
Aaron7 said:
When in equilibrium the springs are length a.
The forces acting in the same direction of motion would be (a/2 - d)k and -(a/2 + d)k since the spring is already stretch by a/2.

I am having trouble finding the vertical forces acting on the mass by the other 2 springs.
The length of the spring would be stretched to √(a^2 + d^2) so the force would be (√(a^2 + d^2) - a/2)k but I need to find the vertical component of this force.
Rather than try to find horizontal and vertical components, turn your axes so that one of then runs along the direction of the displacement. You will then be looking for the components of the spring forces along that axis.

You should be able to pick out a triangle similar to (in the geometric sense) the force triangle from which you want to pick out the appropriate component. You will already have the side lengths for this other triangle.
 
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  • #3
Ok, so with the triangle with sides a and d, I would work out the force in the d direction by saying:
The force in the a direction is k(a/2) so the force in the d direction is k(d/2) because the distances a and d are similar to the forces acting in the a and d directions. This gets the correct answer. Is this valid reasoning?
 
  • #4
Aaron7 said:
Ok, so with the triangle with sides a and d, I would work out the force in the d direction by saying:
The force in the a direction is k(a/2) so the force in the d direction is k(d/2) because the distances a and d are similar to the forces acting in the a and d directions. This gets the correct answer. Is this valid reasoning?

Well, something along those lines, but not exactly that. I don't think you can get away from calculating the change in force due to the extension of the springs. However, if you can find the force f when the spring is in its new position, then the component of that force along the line of the displacement can be had by similar triangles.

If L is the length of the spring at its new location, then the total extension of the spring is L - a/2.

In the figure, the ratio fd/f = d/L holds.

attachment.php?attachmentid=45863&stc=1&d=1333502334.gif


You should be able to determine f from Hook's law and the amount by which L exceeds a/2, the natural length of the spring.
 

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  • #5
So I have f=(L- a/2)k

=> fd = dk(L-a/2) / L
=> fd = dk - dka/2L

Can I then say cosθ = a/L and since d<<L then cosθ = 1 - θ2/2
so a/L = 1 therefore fd = dk/2 and then complete it from there?
 
  • #6
Aaron7 said:
So I have f=(L- a/2)k

=> fd = dk(L-a/2) / L
=> fd = dk - dka/2L
Well that looks okay.
Can I then say cosθ = a/L and since d<<L then cosθ = 1 - θ2/2
so a/L = 1 therefore fd = dk/2 and then complete it from there?

I don't think I'd head off into trig land here. You can find L in terms of a and d, and so fd likewise.

You've got fL as the force along the line of a spring as ##f_L = k\left(\sqrt{a^2 + d^2} - \frac{a}{2}\right)## for one spring. Use similar triangles to find the force fd along the line of displacement. After a bit of simplification you should be able to to see what additional simplification is possible when d << a.
 

FAQ: How Do You Determine the Frequency of a Mass Connected to Four Springs?

What is simple harmonic motion (SHM)?

Simple harmonic motion is a type of periodic motion in which an object oscillates back and forth between two points with a constant amplitude and a fixed period. This type of motion is characterized by a restoring force that is directly proportional to the displacement of the object from its equilibrium position.

How does a mass with 4 springs exhibit SHM?

A mass attached to 4 springs will exhibit SHM when the springs are arranged in a way that they all pull the mass towards the same equilibrium point. As the mass is displaced from its equilibrium position, the springs will exert a restoring force on the mass, causing it to oscillate back and forth in a periodic manner.

What factors affect the SHM of a mass with 4 springs?

The main factors that affect the SHM of a mass with 4 springs are the spring constants of each spring, the mass of the object, and the initial conditions (such as displacement and velocity). These factors determine the amplitude and period of the motion.

How is the period of SHM affected by the mass and spring constants?

The period (T) of SHM is directly proportional to the square root of the mass (m) and inversely proportional to the square root of the spring constant (k). This means that as the mass increases, the period increases, and as the spring constant increases, the period decreases.

Can the SHM of a mass with 4 springs be affected by external forces?

Yes, the SHM of a mass with 4 springs can be affected by external forces such as friction or air resistance. These forces can dampen the oscillations and cause the amplitude to decrease over time. In some cases, external forces can also change the period of the motion.

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