# Identifying Multiple Oscillations in a Graph

• GoCubs12
In summary: I tried these equations and they produce the correct angular frequency if both blocks are initially put at 0.25m to the right of their equilibrium positions. However, the frequency is different if one is put -0.25 from its eq. position. Should this be the case?Yes, the frequency should be different depending on the starting position.
GoCubs12

## Homework Statement

Two masses (M1 and M2) are connected together by 3 springs. The spring constants are k1, k2, k3. Block 1 is at equilibrium at x=0. Block 2 is at equilibrium at x=1. Determine a function for the force on the blocks.

## The Attempt at a Solution

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Without using the initial values, I was able to get the following equations

F1=-k1(x1)+k2(x2-x1)

F2=-k2(x2-x1)-k3(x2)

I looked around the internet and believe these equations are correct, however, my problem is converting them to work with the initial values in the problem. At equilibrium, the force on block 1 is 0 and x1 is 0. Using this I got the equation

F1=-x1(k1+k2)

Doing the same procedure with block 2 where its equilibrium is at 1, I got the following

x1=(k3/k2)+1

so...
F2= -k3(x2)-k2(x2-((k3/k2)+1)

I know the angular frequency of the blocks is about 15.8 rad/s. However, when these equations are used in a program to model the motion of the blocks, the frequency is around 22. I know the program is correct, and so the equations must be wrong, but I have no idea how.

Thanks in advance for the help!

GoCubs12 said:
Two masses (M1 and M2) are connected together by 3 springs.
How are the springs arranged? In series? Parallel? (A diagram would be helpful.)

Doc Al said:
How are the springs arranged? In series? Parallel? (A diagram would be helpful.)

Spring 1 is attached to the left side of block1. Then spring 2 connects the right side of 1 and left the side of 2 together. Then spring 3 goes from the right side of block 2 to another wall. I tried to type out the diagram below.

Wall k1 block 1 k2 block 2 k3 wall

If x1 is the position of block 1, x2 the position of block 2, express the force from each spring in terms of its displacement from equilibrium.

The force from spring 1 is easy since the unstretched position is at x = 0. So it's -k1x1.

When block 2 is at position x2, how much is spring 3 stretched? And then do the same for spring 2.

Doc Al said:
If x1 is the position of block 1, x2 the position of block 2, express the force from each spring in terms of its displacement from equilibrium.

The force from spring 1 is easy since the unstretched position is at x = 0. So it's -k1x1.

When block 2 is at position x2, how much is spring 3 stretched? And then do the same for spring 2.
For spring 3, I got (1-x2) for its displacement
For spring 2, I got (x2-x1)

These still don't seem to be right though

GoCubs12 said:
For spring 3, I got (1-x2) for its displacement
OK. I'd express it as x2-1. When x2 > 1, the spring is compressed.

GoCubs12 said:
For spring 2, I got (x2-x1)
This won't work. At equilibrium, when x2 = 1 and x1 = 0, the amount the spring is stretched is 0, not x2-x1 = 1. Try again.

Note that for simplicity I assume that at equilibrium all springs are at their unstretched length. But that doesn't really matter. The resulting equations will be the same even if you assume some non-zero initial tension, since they are at equilibrium.

Doc Al said:
OK. I'd express it as x2-1. When x2 > 1, the spring is compressed.This won't work. At equilibrium, when x2 = 1 and x1 = 0, the amount the spring is stretched is 0, not x2-x1 = 1. Try again.

Note that for simplicity I assume that at equilibrium all springs are at their unstretched length. But that doesn't really matter. The resulting equations will be the same even if you assume some non-zero initial tension, since they are at equilibrium.
Thanks for the help. So for spring 2 it should be 1-x2-x1?

GoCubs12 said:
So for spring 2 it should be 1-x2-x1?
Almost. The difference in position is x2-x1. When that differs from 1, the spring is stretched or compressed. I'd write that as (x2-x1)-1.

Doc Al said:
Almost. The difference in position is x2-x1. When that differs from 1, the spring is stretched or compressed. I'd write that as (x2-x1)-1.

Okay. So the force on block 1 would be the force from spring 1 and 2.

so... F1=-k1(x1)+(x2-x1-1)k2

and the force on block 2 would be

F2= -k3(x2-1)-(x2-x1-1)K2I tried these equations and they produce the correct angular frequency if both blocks are initially put at 0.25m to the right of their equilibrium positions. However, the frequency is different if one is put -0.25 from its eq. position. Should this be the case? I thought no matter where they start they should always have the same angular frequency as its value is not dependent on the starting positions of the blocks.

GoCubs12 said:
Should this be the case? I thought no matter where they start they should always have the same angular frequency as its value is not dependent on the starting positions of the blocks.
I agree. The frequency of oscillation should not depend on initial positions.

GoCubs12 said:
Block 1 is at equilibrium at x=0. Block 2 is at equilibrium at x=1.
I assume you mean that with both blocks so positioned the system is at equilibrium.
GoCubs12 said:
However, the frequency is different if one is put -0.25 from its eq. position.

So using the forces of motion above I get the following graphs. When x1 is put at an initial position of 0.25 and x2 is at 1.25, the first graph is obtained. When x1 is at -0.25 and x2 is at 1.25, the second graph is obtained. Both times the masses are either in phase or antiphase, which is what I expect, however, the second starting position the period is much faster and I am at a loss to explain it. The first period seems to be what would be expected. I also can't seem to find a way to upload an image of these graphs.

GoCubs12 said:
So using the forces of motion above I get the following graphs. When x1 is put at an initial position of 0.25 and x2 is at 1.25, the first graph is obtained. When x1 is at -0.25 and x2 is at 1.25, the second graph is obtained. Both times the masses are either in phase or antiphase, which is what I expect, however, the second starting position the period is much faster and I am at a loss to explain it. The first period seems to be what would be expected. I also can't seem to find a way to upload an image of these graphs.
How are you creating these graphs - by simulation using the differential equations or by solving them?

haruspex said:
How are you creating these graphs - by simulation?

Yes using MATLAB

haruspex said:
It is feasible that there are two different modes of oscillation at different frequencies, and that in general these would be superimposed. Different initial conditions could change their relative amplitudes.
Good point!

GoCubs12 said:
Could you explain what that means?
As shown at the link I posted, the equations have two solutions for the frequency. In general, both frequencies occur, superimposed. I.e. the displacement of a given mass will be ##A_1\cos(\omega_1t+\phi_1)+A_2\cos(\omega_2t+\phi_2)##. The other mass will have different amplitudes and phases, but the same frequencies.
The values of the phases and amplitudes will depend on the initial conditions. It may be that in one of the two cases you tried one of the two oscillations dominated, and in the other case the other dominated.
Play around with the initial conditions to see if you can observe a case where both oscillations are clearly present.

haruspex said:
As shown at the link I posted, the equations have two solutions for the frequency. In general, both frequencies occur, superimposed. I.e. the displacement of a given mass will be ##A_1\cos(\omega_1t+\phi_1)+A_2\cos(\omega_2t+\phi_2)##. The other mass will have different amplitudes and phases, but the same frequencies.
The values of the phases and amplitudes will depend on the initial conditions. It may be that in one of the two cases you tried one of the two oscillations dominated, and in the other case the other dominated.
Play around with the initial conditions to see if you can observe a case where both oscillations are clearly present.
How would I know when both oscillations are present?

GoCubs12 said:
How would I know when both oscillations are present?
It should be quite clear in the graph. Use Matlab with a sum of two cosine functions, different frequencies but not multiples of each other, similar but different amplitudes. Phase won't matter.

## What is the concept of "3 Springs and 2 Masses"?

The concept of "3 Springs and 2 Masses" refers to a physical system that consists of three springs connected in series with two masses attached to the ends of the outer springs. This system is commonly used in physics experiments to study the principles of simple harmonic motion and the behavior of springs and masses.

## How does the behavior of the system change with different spring constants?

The behavior of the "3 Springs and 2 Masses" system is directly affected by the spring constants of the three springs. The higher the spring constant, the stiffer the spring is and the faster the system will oscillate. Conversely, the lower the spring constant, the more flexible the spring is and the slower the system will oscillate.

## What is the role of the two masses in this system?

The two masses in the "3 Springs and 2 Masses" system serve as the objects that are undergoing simple harmonic motion. They are attached to the ends of the outer springs and their movements are influenced by the spring forces acting on them. The masses also determine the natural frequency of the system.

## How does damping affect the behavior of the system?

Damping is the process of reducing the amplitude of oscillations in a system. In the "3 Springs and 2 Masses" system, damping can be introduced by adding a small amount of friction or resistance to the movement of the masses. This results in a decrease in the amplitude of the oscillations and a change in the system's natural frequency.

## What is the equation that describes the motion of the system?

The motion of the "3 Springs and 2 Masses" system can be described by the equation: x(t) = A cos(ωt + φ), where x(t) is the position of the mass at time t, A is the amplitude of the oscillations, ω is the angular frequency, and φ is the phase angle. This equation is derived from the principles of simple harmonic motion and can be used to analyze the behavior of the system.

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