Solve 2 Mass Spring Oscillations: m1,m2,k

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In summary, the given two-body problem of two masses connected by a spring can be reduced to a one-body problem by choosing the position of the centre of mass and the change of length of the spring as new coordinates. The resulting Lagrangian can be solved to obtain the equations of motion, which show that the centre of mass stays at rest or travels with its original velocity, while the variable u performs a simple harmonic motion with an angular frequency of sqrt(k/mu). This solution can then be used to obtain the solution of the original problem.
  • #1
Cosmossos
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Homework Statement


A spring of constant k connects two masses, m1 and m2. The system is constrained
to longitudinal oscillations (in the direction of the line connecting the masses).
(a) Find the equivalent 1D problem and solve the equations of motion.
(b) Use the solution to the previous point to obtain the solution of the original
problem.

Homework Equations


I solved point a. I got that the lagrangian is :(0.5*m1*v1^2)+0.5*m2*v2^2)-0.5*k*(x1+x2-lo)
Is that correct?
And the solution of the NDE are without cos,sin because x1''=-k/m1 for example.
was I wrong?

Now for point b, I should you coordinate x1,y1,x2,y2? because the masses can also go up and down during their motion?
How will the lagrangian will look like?
Please help me.
 
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  • #2
Now I could be wrong on this, but I would think an equivalent 1D problem would be a simple harmonic oscillator. If you place the origin (and your reference point) on the left mass, it will only appear that the right mass is moving. The Lagrangian for that would be

[tex]
L=\frac{1}{2}m\dot{x}^2+\frac{1}{2}kx^2
[/tex]

because you only have 1 mass moving with 1 direction, hence no subscripts to distinguish particles.

For solving this actual problem, you won't need to consider the [itex]y[/itex] axis at all because you are told the motion is longitudinal only. What you should do is consider the unstretched length of the spring, call it [itex]\ell[/itex] and then use some reference point that is some distance, call it [itex]x_0[/itex] to the left of the left mass. Hopefully this little picture can help describe what I'm talking about:

..|....0/\/\/\/\/\/\/\/\0
..|...[itex]x_0[/itex]...|...[itex]\ell[/itex]...|
 
  • #3
But from the question , How can I be sure that only one mass is moving and not both of them? and regrading to the actual problem , in the 1D problem I also have some stretched length of the the spring , so What's the difference between the two problem?
 
  • #4
If we call the left mass [itex]m_1[/itex] and the second mass is [itex]m_2[/itex], then when we place our origin on [itex]m_1[/itex], its position is [itex]x_{m_1}=0[/itex]; its time-derivative would also be zero: [itex]\dot{x}_{m_1}=d(x_{m_1})/dt=d(0)/dt=0[/itex].

In the second scenario, your origin is not at [itex]m_1[/itex], but somewhere to the left of it. So then its position would be [itex]x_{m_1}=x_0+ x_1[/itex] so that its time-derivative would be [itex]d(x_{m_1})/dt=dx_1/dt=\dot{x}_1[/itex] because [itex]x_0[/itex] is a constant. Your Lagrangian in this case would then be something like

[tex]
L=\frac{1}{2}m_1\dot{x}_1^2+\frac{1}{2}m_2\dot{x}_2^2+\frac{1}{2}k(x_0+x_1)^2+\frac{1}{2}k(x_0+\ell+x_2)^2
[/tex]

As you can see, there is a difference in the Lagrangian, so there will be a difference in equations of motion. You should find http://en.wikipedia.org/wiki/Normal_mode" of oscillation with this problem (I think there will be only 2 modes of oscillation with this set up).
 
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  • #6
Now another thing I don't understand is why you took the 1D problem as the simple harmonic oscillator?
 
  • #7
Cosmossos said:
Now another thing I don't understand is why you took the 1D problem as the simple harmonic oscillator?

If you let [itex]x_0=x_1=0[/itex], then the Lagrangian I wrote in post #4 reduces to the Lagrangian in post #2
 
  • #8
and still, the solution to the motion equations is with exp, and not sin and cos... and also not imaginary soultion
 
  • #9
Cosmossos said:
and still, the solution to the motion equations is with exp, and not sin and cos... and also not imaginary soultion

Whether or not an imaginary solution exists depends on how the potential is defined. But you should recognize that, through the Euler relations,

[tex]
\exp\left[\pm ix\right]=\cos\left[x\right]\pm i\sin\left[x\right]
[/tex]

So either the exponential or the sine/cosine answer is correct. You would only be able to definitively decided which would give the right answer when you have initial conditions.
 
  • #11
Cosmossos said:

Homework Statement


A spring of constant k connects two masses, m1 and m2. The system is constrained
to longitudinal oscillations (in the direction of the line connecting the masses).
(a) Find the equivalent 1D problem and solve the equations of motion.
(b) Use the solution to the previous point to obtain the solution of the original
problem.

This is a two-body problem that can be reduced to one-body problem by appropriate choice of coordinates. As this is a closed system, the centre of mass does nor accelerate. So we choose Xc, the position of centre of mass as one of the new coordinates.
The potential energy of the system depends on the change of length of the spring. The spring connects the two masses, their distance is equal to the length of the spring, x2-x1=L. The potential energy is PE=1/2k(L-L0)2. It is obvious to choose x2-x1-L0 = u as the other new coordinate.

Now we have to write the KE with the new variables. It is easy to show that

[tex]x_1=X_c-\frac{m_2}{m_1+m_2}L[/tex]

[tex]x_2=X_c+\frac{m_1}{m_1+m_2}L[/tex]

The time derivatives are:

[tex]\dot x_1=\dot X_c-\frac{m_2}{m_1+m_2}\dot u [/tex]

[tex]\dot x_2=\dot X_c+\frac{m_1}{m_1+m_2}\dot u [/tex]

The expression for the KE can be obtained in terms of the time derivatives of the new variables:

[tex]KE=\frac{1}{2} (m_1\dot x_1^2+m2 \dot x_2^2)= \frac{1}{2} (m_1+m_2)\dot X_c^2+\frac{1}{2} \frac{m_1*m_2}{m_1+m_2} \dot u^2
[/tex]

[tex]\frac{m_1*m_2}{m_1+m_2}=\mu [/tex]

is called the reduced mass of the system.

The Lagrangian of the system :

[tex]L=KE-PE = \frac{1}{2} (m_1+m_2)\dot X_c^2+\frac{1}{2} *\mu \dot u^2 -\frac{1}{2}k* u^2
[/tex]

The Lagrange equations of motion:

[tex](m1+m2)\frac{d^2 X_c}{dt^2}=0 [/tex]

the centre of mass stays in rest or travels with its original velocity,

[tex]\mu\frac{d^2 u}{dt^2}+k*u=0 [/tex]

the variable u performs a simple harmonic motion [tex]u=A\sin(\omega t +\alpha) [/tex]

with the angular frequency

[tex] \omega=\sqrt{\frac{k}{\mu}}[/tex]

Inserting Xc and u back into the expression for x1 and x2, we obtain the solution of the original problem: The two masses will oscillate around the CM with the same frequency and phase.

ehild
 

Related to Solve 2 Mass Spring Oscillations: m1,m2,k

1. What is the equation for mass spring oscillations?

The equation for mass spring oscillations is:
x(t) = A cos(ωt + φ)
Where,
x(t) = displacement of the mass from equilibrium
A = amplitude of oscillation
ω = angular frequency
φ = phase angle

2. How do you calculate the angular frequency of a mass spring system?

The angular frequency (ω) of a mass spring system can be calculated using the formula:
ω = √(k/m)
Where,
k = spring constant
m = mass of the object

3. What is the relationship between the mass and the angular frequency in a mass spring system?

In a mass spring system, the angular frequency (ω) is inversely proportional to the mass (m). This means that as the mass increases, the angular frequency decreases and vice versa. This relationship is represented by the formula:
ω ∝ 1/m

4. How do you solve for the displacement of each mass in a dual mass spring system?

To solve for the displacement of each mass in a dual mass spring system, you will need to use the equations of motion. These equations are:
m1x1'' + k(x1-x2) = 0
m2x2'' + k(x2-x1) = 0
Where,
x1 and x2 are the displacements of mass m1 and m2 respectively.

5. Can the mass spring system exhibit simple harmonic motion?

Yes, a mass spring system can exhibit simple harmonic motion as long as the restoring force is directly proportional to the displacement from equilibrium. This means that the system must follow Hooke's Law, where the force is equal to the negative of the spring constant multiplied by the displacement. This results in a sinusoidal motion, which is a characteristic of simple harmonic motion.

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