Coupled oscillators -- period of normal modes

In summary, the conversation discusses the existence of two normal modes in a system with two masses and the use of Newton's law to find the equations for each mode. The pendulum mode is when the masses have the same displacement while the breathing mode is when they have opposite displacements. The periods for each mode are determined by the equations ##\omega_p = \sqrt{\frac{g}{l}}## and ##\omega_b = \sqrt{\frac{g}{l} + \frac{2k}{m}}##, with the latter being more complicated due to the variables k and m. The conversation also mentions finding the period for the normal modes and the potential use of the expressions ##\frac{1.5}{2
  • #1
happyparticle
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21
Homework Statement
I have to determine the periods of the normal modes.

I have 2 mass ##m_1 = m_2##, l = 1.2m and if I hold vertically one of the mass the period of the other mass is 1.5s.
Relevant Equations
F = ma

##m\ddot x = -\frac{mgx_1}{l} -k(x_1 - x_2)##
##m\ddot x_2 = -\frac{mgx_2}{l} +k(x_1 - x_2)##
Hi,

I know there's are 2 normal modes because the system has 2 mass. I did the Newton's law for both mass.

##m\ddot x_1 = -\frac{mgx_1}{l} -k(x_1 - x_2)## (1)
##m\ddot x_2 = -\frac{mgx_2}{l} +k(x_1 - x_2)## (2)

In the pendulum mode ##x_1 = x_2## and in the breathing mode ##x_1 = -x_2##

I get the pendulum and breathing mode by adding equation 1 and 2 and subtracting 1 and 2 then I replace ##x_1-x_2## by q2 and ##x_1+x_2## by q1.
Finally I have ##\omega_p = \sqrt{\frac{g}{l}}## and ##\omega_b = \sqrt{\frac{g}{l} + \frac{2k}{m}}##
It's quite easy to get the period for ##\omega_p##. However, I'm not sure how to find the period for ##\omega_b##, since I don't have k and m.

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  • #2
You will need to supply a diagram or a full description of the physical arrangement. In particular, I am confused regarding whether these x displacements are horizontal or vertical.
 
  • #3
haruspex said:
You will need to supply a diagram or a full description of the physical arrangement. In particular, I am confused regarding whether these x displacements are horizontal or vertical.
My bad, sorry. I uploaded an image.
 
  • #4
EpselonZero said:
My bad, sorry. I uploaded an image.
Doesn't this tell you k/m?
"if I hold vertically one of the mass the period of the other mass is 1.5s."
 
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  • #5
haruspex said:
Doesn't this tell you k/m?
"if I hold vertically one of the mass the period of the other mass is 1.5s."

I need to find the period for the normal modes. I don't see the link between 1.5s and the normal modes.
 
  • #6
EpselonZero said:
I need to find the period for the normal modes. I don't see the link between 1.5s and the normal modes.
You wrote that what was blocking you was that you did not know k/m.
What wouid the period be if you suspended one of the masses from the other and let it oscillate vertically?
 
  • #7
I see...

Can I plug ##\frac{1.5}{2\pi}## to find ##\omega_b##?

It seems too easy.
 
  • #8
EpselonZero said:
I see...

Can I plug ##\frac{1.5}{2\pi}## to find ##\omega_b##?

It seems too easy.
not quite that simple... it tells you k/m. Your equation for ##\omega_b## has another term
 
  • #9
##\omega_b = \sqrt{\frac{g}{l} + \frac{2k}{m}}##
##g = 9.8## and ##l = 1.2##

##\omega_b = \sqrt{8.2 + 2(\frac{1.5}{2pi})}##

Is it correct?
 
  • #10
EpselonZero said:
##\omega_b = \sqrt{\frac{g}{l} + \frac{2k}{m}}##
##g = 9.8## and ##l = 1.2##

##\omega_b = \sqrt{8.2 + 2(\frac{1.5}{2pi})}##

Is it correct?
Doesn’t look right to me.
What is the expression for the period of a mass on a spring?
 
  • #11
Right, I made a mistake.

##\omega_b = \sqrt{8.2 +2(\frac{2\pi}{1.5})^2}##
 
  • #12
EpselonZero said:
Right, I made a mistake.

##\omega_b = \sqrt{8.2 +2(\frac{2\pi}{1.5})^2}##
Right.
 
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FAQ: Coupled oscillators -- period of normal modes

What are coupled oscillators?

Coupled oscillators are a system of two or more oscillators that are connected or coupled in some way, such as through a spring or pendulum. The motion of one oscillator affects the motion of the other oscillator, resulting in complex patterns of oscillation.

What is the period of normal modes in coupled oscillators?

The period of normal modes in coupled oscillators refers to the time it takes for the system to complete one full cycle of oscillation. This period is dependent on the mass, stiffness, and coupling strength of the oscillators.

How is the period of normal modes calculated?

The period of normal modes can be calculated using the formula T = 2π√(m/k), where T is the period, m is the mass of the oscillators, and k is the spring constant or stiffness of the system.

What are the normal modes of coupled oscillators?

The normal modes of coupled oscillators are the natural modes of oscillation that the system exhibits. These modes are characterized by specific frequencies and patterns of motion, and can be visualized as standing waves within the system.

What factors affect the period of normal modes in coupled oscillators?

The period of normal modes in coupled oscillators is affected by the masses and stiffness of the oscillators, as well as the strength of the coupling between them. Additionally, external forces and damping can also affect the period of normal modes.

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