Short Answer SHM of diatomic molecule

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SUMMARY

The discussion focuses on the harmonic motion of diatomic molecules, specifically addressing the momenta and kinetic energy of two identical atoms vibrating as harmonic oscillators. It establishes that the momenta of the atoms relative to their center of mass are p and -p, and that the total kinetic energy K can be represented as that of a single object with mass m/2 and momentum p. Additionally, it confirms that for non-identical atoms with masses m1 and m2, the concept of reduced mass, defined as (m1)(m2)/(m1+m2), applies, maintaining the validity of the previous results.

PREREQUISITES
  • Understanding of harmonic oscillators in physics
  • Familiarity with the concept of center of mass
  • Knowledge of kinetic energy equations
  • Basic principles of momentum conservation
NEXT STEPS
  • Study the derivation of reduced mass in multi-body systems
  • Explore the implications of harmonic motion in molecular vibrations
  • Learn about the mathematical formulation of kinetic energy in oscillatory systems
  • Investigate the relationship between momentum and forces in isolated systems
USEFUL FOR

Students studying classical mechanics, particularly those focusing on molecular physics, as well as educators seeking to clarify concepts related to harmonic oscillators and momentum conservation.

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Homework Statement



Two identical atoms in a diatomic molecule vibrate as harmonic oscillators. However, their center of mass, midway between them, remains at rest.

1)Show that at any instant, the momenta of the atoms relative to the center of mass are p and -p

2)Show that the total kinetic energy K of the two atoms at any instant is the same as that of a single object with mass m/2 with a momentum of magnitude p .

3)If the atoms are not identical but have masses m_1 and m_2, show that the result of part (a) still holds and the single object's mass in part (b) is (m_1)(m_2)/(m_1+m_2). The quantity (m_1)(m_2)/(m_1+m_2) is called the reduced mass of the system.

Homework Equations



Energy= 1/2mv^2 + 1/2kx^2 = 1/2kA^2
Momentum= p= mv
Period= T = 2pi (m/k)^(1/2)

The Attempt at a Solution



This is a short answer problem and I really don't know how to go about this for sure. I think that you can assume that it is working like a spring.

1) I don't know how to show any work for number one, but would it be p= -p because there are no external forces acting? They are the same atoms, so they have to have the same mass and velocity and p=mv.

2)For this all I think is that the KE of the single object would be:
1/2(m/2)v^2= 1/2kA^2. This is for the case when x=0, so there is no potential energy from the spring.

3)This I have no idea.

We really haven't been over anything like this in class, so I am not sure what to do. Thanks for any help.
 
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