# Coupled spring pendulum (harmonic oscillation)

1. Mar 29, 2016

### JulienB

1. The problem statement, all variables and given/known data

Hi everybody!

Two masses m1 and m2 are connected with a spring one after the other to a wall (see attached picture). The spring constants are k1 and k2. To consider here are only longitudinal oscillations and no external forces.
a) Express the newtonian equations of motion, when x1 and x2 represent the deflections of the two masses.
b) Calculate the eigenfrequency ωi of the coupled system for the case m1 = m2 = m and k1 = k2 = k. A possible approach is x1 = A⋅cos(ωt) and x2 = B⋅cos(ωt). An explicit isolation of the equations of oscillation is here not necessary.

2. Relevant equations

spring forces, differential equations for motion

3. The attempt at a solution

Okay I'm still pretty confused about such oscillations (and unfortunately differential equations as well ), but I gave it a go by trying to determine the acceleration for each mass:

a)
m1⋅a1 = FF1 - FF12 (see picture)
⇔ m1⋅d2x1/dt2 = -k1x1 + k2(x2 - x1)
⇔ m1⋅d2x1/dt2 + k1⋅x1 - k2⋅x2 + k2⋅x1 = 0

m2⋅a2 = FF21 (see picture)
⇔ m2⋅d2x2/dt2 = - k2(x2 - x1)
⇔ m2⋅d2x2/dt2 + k2⋅x2 - k2⋅x1 = 0

Is that correct? Am I to do anything more to those equations or is that a sufficient answer to a)?

b)
Here I don't really know what I should be doing... I coupled the equations from a) together and applied m1 = m2 = m and k1 = k2 = k:

m⋅d2x1/dt2 + m⋅d2x2/dt2 + k⋅x1 - k⋅x2 + k⋅x1 + k⋅x2 - k⋅x1 = 0
⇔ m⋅(d2x1/dt2 + d2x2/dt2) + k⋅x1 = 0

Then I substituted d2x1/2/dt2 with their respective second derivatives:

m⋅(-A⋅ω2⋅cos(ω⋅t) - B⋅ω2⋅cos(ωt)) + k⋅(A⋅cos(ωt)) = 0
⇔ m⋅ω2⋅(A + B) = k⋅A
⇔ ω2 = k⋅A/m⋅(A + B)
⇔ ω = √(k⋅A/m⋅(A+B))

That's something... But to be honest I have almost no idea what I am doing, I just try to progress in the direction of what the problem asks me. Does ωi mean there is a different frequency for each mass? Should I be able to substitute A and B with something else?

Julien.

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2. Mar 29, 2016

### ehild

Yes, it is correct, and a sufficient answer to a).
You do not need to couple the equations at the beginning, just substitute the trial solution x1=Acos(wt), x2=Bcos(wt). You get a homogeneous linear system of equations for A and B which has nonzero solution at certain w-s. These are the two eigenfrequencies of the system. Both masses move with these angular frequencies when performing a "normal mode". The general motion is a linear combination of these normal modes.

3. Mar 29, 2016

### TSny

Hi, Julien!

For part (b), I think you just need to substitute the expressions x1 = Acosωt and x2 = Bcosωt into your equations from part(a) and solve for ω.

4. Mar 29, 2016

### JulienB

@ehild @TSny Thank you both for your answers, it's very reassuring to have something correct!

Unfortunately b) is still not working so well. I didn't mix the two equations and labeled ω1 and ω2, but this turned into a nightmare (see attached picture). I didn't go further yet, because I guess I'm not using the right method. Any advice?

Thank you so much.

Julien.

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5. Mar 29, 2016

### JulienB

Am I allowed to replace the k/m in my last expressions by ω22?

6. Mar 29, 2016

### TSny

For convenience, you can let k/m be represented by ωo2, say. But ωo does not represent the angular frequency of either mass when the system is oscillating in a normal mode. For a normal mode, each mass oscillates with the same angular frequency ω which you want to determine.

After simplifying your equations from part (a) for the special case of equal masses and spring constants, substitute the given expressions for x1 and x2 and solve for ω in terms of ωo.

7. Mar 29, 2016

### TSny

You have two equations from part (a). After substituting the given expressions for x1 and x2, you should have two equations involving A, B, and ω. But, try to write these in terms of just ω and the ratio A/B. You then have two equations for two unknowns: ω and A/B.

8. Mar 29, 2016

### JulienB

@TSny Thank you again for your answer (and a special thanks for your help on all my other posts as well :) ). I will give it some thought and trial and come back tomorrow with a new attempt at solving it.

Julien.

9. Mar 29, 2016

### ehild

Do not mix terms with ω1 and ω2. Both masses move with the same frequency. Substitute just x1=Acos(ωt) and x2=Bcos(ωt) into the differential equations. After differentiation, you can simplify the equations by cos(ωt), and you get two equations for A, B, and ω. You get two solutions, with two values for ω.

10. Mar 30, 2016

### JulienB

Okay I come back with a new solution. I think it is correct now, but there is still a few things I'd like to clarify for myself:

So by substituting x1 and x2 inside my equations from a), I get:

1) -m⋅A⋅ω2⋅cos(ω1⋅t) + 2⋅k⋅A⋅cos(ω1⋅t) - k⋅B⋅cos(ω2⋅t) = 0
2) -m⋅B⋅ω2⋅cos(ω2⋅t) + k⋅B⋅cos(ω2⋅t) - k⋅A⋅cos(ω1⋅t) = 0

Then I wondered what would the equation show me at t=0:

1') -m⋅A⋅ω2 + 2⋅k⋅A - k⋅B = 0 ⇔ A(2⋅k - m⋅ω2) = k⋅B
2') -m⋅B⋅ω2 + k⋅B - k⋅A = 0 ⇔ A = B(k - m⋅ω2)/k

Am I allowed to do that? I'm not sure how to consider ω, ω1, ω2... Can someone explain what each one represents in the system? Like why can I use ω without subscript? Because now I don't know if they cancel because t = 0 or if because cos(ω1⋅t) is to be considered equal to ω2...

Anyway, when I input 2') in 1'), I get:

(k - m⋅ω2)⋅(2k - m⋅ω2) = k2

Solving for ω I get ω1/2 = √(((3 ± √5)/2)⋅(k/m)). Does that make sense?

Thank you very much for your help, I'm getting there!

Julien.

11. Mar 30, 2016

### ehild

You substituted wrong expressions for x1 and x2, and your second derivatives are also wrong. the second derivative of cos(ω1t) is
12cos(ω1t),
and that of cos(ω2t) is
22cos(ω2t).
Both objects move with the same frequency. x1=A cos(ωt) and x2=Bcos(ωt). Substitute these expressions into the differential equation.

12. Mar 30, 2016

### JulienB

@ehild Hi and thanks for your answer. I don't get it, you mean I should not have ignored the subscript on the ω or that the derivative is completely wrong? The second derivative of A⋅cos(ω1⋅t) is -A⋅ω12⋅cos(ω1⋅t) right?

13. Mar 30, 2016

### haruspex

There is an ω1 and ω2 because there are two frequencies at which the system can oscillate 'nicely'. But in each case, each mass oscillates at that common frequency. So handle all the equations with just ω, and you should end up with a quadratic for its value.

14. Mar 30, 2016

### ehild

Yes, your derivatives are completely wrong.What is the second derivative of cos(3t), for example? is it not -9cos(3t) ?
You should find solutions of the system of differential equations where all objects move with the same ω. All replies said that already.
You get two ω-s, that means, the problem has two solutions, two different vibrations.These vibrations are called the normal modes of the system, and the general solution is a linear combination of these normal modes. You will learn about it later.

Last edited: Mar 30, 2016
15. Mar 31, 2016

### JulienB

@haruspex @ehild Thank you for your answers. I'll try to make the best of it.

That's very sad! And I still don't exactly get why I'm afraid...Let's start b) all over again with the solutions given in the problem:

x1 = A⋅cos(ωt) and x2 = B⋅cos(ωt)

I take their first and second derivatives:

x1' = -A⋅ω⋅sin(ωt) and x2' = -B⋅ω⋅sin(ωt)
x1'' = -A⋅ω2⋅cos(ωt) and x2'' = -B⋅ω2⋅cos(ωt)

The only difference with what I've done in my last post is to take ω instead of ω1/2, like haruspex said.

Then I input those in the equations I got in a):

1) -m⋅A⋅ω2⋅cos(ωt) + k⋅A⋅cos(ωt) - k⋅B⋅cos(ωt) + k⋅A⋅cos(ωt) = 0
2) -m⋅B⋅ω2⋅cos(ωt) + k⋅B⋅cos(ωt) - k⋅A⋅cos(ωt) = 0

from which I get:

1') (2⋅k - m⋅ω2)⋅A = k⋅B
2') A = (k - m⋅ω2)⋅B/k

I input 2') in 1'):

(2⋅k - m⋅ω2)⋅(k - m⋅ω2)⋅B/k = k⋅B

I remove the B, multiply by k on both sides and I get:

ω4 - 3⋅(k/m)⋅ω2 + k2/m2 = 0

I solve using the quadratic formula but get the same result as before:

ω2 = (3/2)(k/m) ± √((9/4)(k2/m2) - (k2/m2))
ω1 = √((3 + √5)k/(2n) and ω2 = √((3 - √5)k/(2n)

It's the same strange looking result I got in my previous post, only I used one frequence for both equations as you advised me and solved for ω. Is that the right way to deal with it? I guess by practice I'll visualize it better and better.

Julien.

16. Mar 31, 2016

### ehild

There is only a typo now (n instead of m). $$\omega_1=\sqrt{\frac {k}{2m} (3+\sqrt{5})}$$
$$\omega_2=\sqrt{\frac {k}{2m} (3-\sqrt{5})}$$

17. Mar 31, 2016

### JulienB

@ehild Alright, thank you very much for all your help and sorry for getting it slowly :) thank you @haruspex too!