Coupled spring pendulum (harmonic oscillation)

[JulienB]

Homework Statement

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 ω_i 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.

Homework Equations

spring forces, differential equations for motion

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 = F_F1 - F_F12 (see picture)
⇔ m₁⋅d²x₁/dt² = -k₁x₁ + k₂(x₂ - x₁)
⇔ m₁⋅d²x₁/dt² + k₁⋅x₁ - k₂⋅x₂ + k₂⋅x₁ = 0

m₂⋅a₂ = F_F21 (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 d²x_1/2/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 ω_i mean there is a different frequency for each mass? Should I be able to substitute A and B with something else?Thank you very much for your answers, I appreciate it.Julien.

 

[ehild]

Homework Statement

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 ω_i 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.

Homework Equations

spring forces, differential equations for motion

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 = F_F1 - F_F12 (see picture)
⇔ m₁⋅d²x₁/dt² = -k₁x₁ + k₂(x₂ - x₁)
⇔ m₁⋅d²x₁/dt² + k₁⋅x₁ - k₂⋅x₂ + k₂⋅x₁ = 0

m₂⋅a₂ = F_F21 (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)?

Yes, it is correct, and 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 d²x_1/2/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 ω_i mean there is a different frequency for each mass? Should I be able to substitute A and B with something else?Thank you very much for your answers, I appreciate it.Julien.

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.

 

[TSny]

Hi, Julien!

Your answer for part (a) looks good to me.

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.

 

[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.

 

[JulienB]

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

 

[TSny]

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

For convenience, you can let k/m be represented by ω_o², say. But ω_o 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 ω_o.

 

[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.

 

[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.

 

[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.

 

[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 ω_1/2 = √(((3 ± √5)/2)⋅(k/m)). Does that make sense?

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

 

[ehild]

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

You substituted wrong expressions for x1 and x2, and your second derivatives are also wrong. the second derivative of cos(ω₁t) is
-ω₁²cos(ω₁t),
and that of cos(ω₂t) is
-ω₂²cos(ω₂t).
Both objects move with the same frequency. x₁=A cos(ωt) and x₂=Bcos(ωt). Substitute these expressions into the differential equation.

 

[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?

 

[haruspex]

@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?

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.

 

[ehild]

@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?

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.

 

[JulienB]

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

Yes, your derivatives are completely wrong.

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 ω_1/2, 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.

 

[ehild]

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

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..

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

 

[JulienB]

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