Normal modes, 2 masses, 2 springs

In summary, the conversation was about a normal modes question involving a spring suspended from a ceiling with two masses and two spring constants. The problem was to calculate the normal mode frequencies, denoted as alpha, and the attempted solution involved using the displacement of the two masses and a determinant equation to solve for alpha. The conversation also mentioned difficulties with using the equation writer on the forum.
  • #1
Hume Howe
1
0

Homework Statement


hey, this is my first time using this forum and was wondering if i could have some help with this normal modes question.

suspended from a ceiling is in order: a spring of spring constant ([tex]k_{2}[/tex]), a mass (m), a spring with spring constant ([tex]k_{1}[/tex]), and another mass (m)

calulate the normal mode frequencies [tex]\alpha_{1}[/tex], [tex]\alpha_{1}[/tex] sorry can't find the omega symbol.

attempted solution:
let the displacement of the higher mass be x and the displacement of the lower mass be y.

m[tex]\ddot{x}[/tex]=-[tex]k_{2}[/tex]x + [tex]k_{1}[/tex](y-x)
m[tex]\ddot{y}[/tex]=-[tex]k_{1}[/tex](y-x)

[tex]\left| \alpha^{2} - (\frac{k_{2}}{m} + \frac{k_{1}}{m}) -----\frac{k_{1}}{m} \right|[/tex]
[tex]\left| \frac{k_{1}}{m} -------- \alpha^{2} - \frac{k_{1}}{m} \right|[/tex] = 0

determinant = 0

so [tex]\alpha^{2}[/tex] = 1/2[ ([tex]\frac{k_{2}}{m}[/tex]+ [tex]\frac{2k_{1}}{m}[/tex]) +- sqrt[([tex]\frac{k_{2}}{m}[/tex]+ [tex]\frac{2k_{1}}{m}[/tex])^2 – 4([tex]\frac{k_{2}*k_{1}}{m^2}[/tex])]]

after simplifying it doesn't work when I use to try and solve for the amplitude ratios.
please advise, thanks.
p.s. is there a simpler equation writer i could download and use here or should i just persevere and try to use this embedded one?
 
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  • #2
Homework Equations m*\ddot{x} = -k_{2}*x + k_{1}(y-x)m*\ddot{y} = -k_{1}(y-x)The Attempt at a Solution let the displacement of the higher mass be x and the displacement of the lower mass be y.m\ddot{x}=-k_{2}x + k_{1}(y-x)m\ddot{y}=-k_{1}(y-x)\left| \alpha^{2} - (\frac{k_{2}}{m} + \frac{k_{1}}{m}) -----\frac{k_{1}}{m} \right|\left| \frac{k_{1}}{m} -------- \alpha^{2} - \frac{k_{1}}{m} \right| = 0determinant = 0so \alpha^{2} = 1/2[ (\frac{k_{2}}{m}+ \frac{2k_{1}}{m}) +- sqrt[(\frac{k_{2}}{m}+ \frac{2k_{1}}{m})^2 – 4(\frac{k_{2}*k_{1}}{m^2})]]\alpha_{1,2} = 1/2[ (\frac{k_{2}}{m}+ \frac{2k_{1}}{m}) +- sqrt[(\frac{k_{2}}{m}+ \frac{2k_{1}}{m})^2 – 4(\frac{k_{2}*k_{1}}{m^2})]]^{\frac{1}{2}}
 
  • #3



Hello, welcome to the forum! It looks like you are working on a problem involving normal modes with two masses and two springs. This is a common problem in physics and it's great that you are seeking help. Let me help you with your attempted solution and provide some guidance to help you solve this problem.

First, it is important to note that there are two normal modes in this system, one where both masses move in phase (in the same direction) and one where they move out of phase (in opposite directions). These two normal modes will have different frequencies and amplitudes.

To solve for the normal mode frequencies, we can use the equations you have set up for the displacements of the two masses. However, there is a small error in your equations. The second equation should be m\ddot{y}=-k_{1}(y-x), not m\ddot{y}=-k_{1}(y-x). This is because the spring with spring constant k_{1} is attached to both masses, so the force it exerts on each mass will be equal in magnitude but opposite in direction.

Now, to find the normal mode frequencies, we can use the equations of motion for each mass and solve for \ddot{x} and \ddot{y}. Substituting these values into the determinant you have set up, we can solve for the normal mode frequencies.

As for the amplitude ratios, we can use the equations for the displacements of the two masses and solve for the ratio of x to y. This will give us the amplitude ratio for each normal mode.

As for the equation writer, you can download a free one such as LaTeX or MathType to make writing equations easier. However, it is good to practice using the embedded one as well.

I hope this helps and good luck with your problem!
 

Related to Normal modes, 2 masses, 2 springs

1. What are normal modes in the context of 2 masses and 2 springs?

Normal modes refer to the oscillation patterns that occur when two masses connected by two springs are displaced from their equilibrium positions. These modes represent the natural frequencies of the system and are characterized by the amplitudes and phases of the masses' movements.

2. How do the masses and springs affect the normal modes?

The masses and springs in the system determine the natural frequencies of the normal modes. The mass of each object affects the inertia of the system, while the stiffness of the springs determines how easily the masses can oscillate. A change in either of these parameters can alter the normal modes of the system.

3. Can the masses and springs have different values and still produce the same normal modes?

Yes, it is possible for different combinations of masses and springs to produce the same normal modes. This is because the natural frequencies of the system are determined by the ratio of the mass and spring constants, rather than their individual values.

4. How can normal modes be visualized in a 2-mass, 2-spring system?

Normal modes can be visualized by plotting the displacement of each mass as a function of time. This will reveal the characteristic oscillation patterns of the system at its natural frequencies. Additionally, normal modes can be represented as standing waves, where the nodes and antinodes correspond to the equilibrium and maximum displacement points of the masses.

5. What is the significance of normal modes in 2-mass, 2-spring systems?

Normal modes are important in understanding the behavior of mechanical systems, such as bridges and buildings, as well as in other fields such as acoustics and electronics. They allow us to predict and control the frequencies at which a system will vibrate, which is crucial for ensuring stability and avoiding resonance.

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