Quantum Number of a Harmonic Oscillator State

[quietrain]

Homework Statement

Consider a harmonic oscillator with mass=0.1kg, k=50N/m , h-bar=1.055x10^-34
Let this oscillator have the same energy as a mass on a spring, with the same k and m, released from rest at a displacement of 5.00 cm from equilibrium. What is the quantum number n of the state of the harmonic oscillator?

The Attempt at a Solution

ok, so i was wondering, why can't i use angular momentum quantization?

meaning mvr = n(h-bar)
so i divide both sides by r²
i get m(k/m)^-1/2=n(h-bar)/r²

make n subject, sub in all values, i get 5.29*10³¹.. which happens to be twice the correct answer.

so did i make a careless mistake somewhere? or was i never been able to use this formula? thanks!

 

[ehild]

Before applying a formula, think about the meaning of the symbols. mvr is the angular momentum of a point mass orbiting along a circle of radius r. But here the body oscillates along a straight line.

The energy of a quantum harmonic oscillator is

$$E=\hbar \omega (v+1/2)$$

v is the vibrational quantum number, a non-negative integer.

ehild

 

[quietrain]

hmm... so you mean that using my attempt's formula is totally wrong in this case?

but is there a reason why i get exactly twice the correct answer?

is it coincidence? or is there some underlying principle?

 

[ehild]

A circular motion can be considered as two perpendicular harmonic oscillators.

 

[quietrain]

oh. but why?

what is the quantity that relates them both?

is it the distance? the speed? the momentum? or?

 

[ehild]

Let a point mass vibrate along the x direction according to the formula x=A cos (wt), and at the same time it moves and along the y direction: y=A sin(wt). The resultant trajectory is a circle of radius A,
and the body moves with angular velocity w and linear velocity v=Aw. The angular momentum is

$$A^2\omega m=\hbar n$$.

The energy of the circular motion is the sum of the kinetic energy and the potential energy, and it is the same as twice the energy of a single linear oscillator.

$$0.5v^2 m+0.5 kA^2 = m A^2 \omega ^2=\hbar \omega (n+0.5)$$

n is so big that the 1/2 term is negligible, so you have the same quantum numbers from the energy of two harmonic oscillators as from the angular momentum.

ehild

 

[quietrain]

Let a point mass vibrate along the x direction according to the formula x=A cos (wt), and at the same time it moves and along the y direction: y=A sin(wt). The resultant trajectory is a circle of radius A,
and the body moves with angular velocity w and linear velocity v=Aw. The angular momentum is

$$A^2\omega m=\hbar n$$.

The energy of the circular motion is the sum of the kinetic energy and the potential energy, and it is the same as twice the energy of a single linear oscillator.

$$0.5v^2 m+0.5 kA^2 = m A^2 \omega ^2=\hbar \omega (n+0.5)$$

n is so big that the 1/2 term is negligible, so you have the same quantum numbers from the energy of two harmonic oscillators as from the angular momentum.

ehild

oh, so is it right to say that KE of 1 linear oscillator = 1/2 mv^2 ?so we assume KE of single linear oscillator = PE of it. so we have 2 times?

so total energy of circular motion = total energy of oscillator?

 

[ehild]

oh, so is it right to say that KE of 1 linear oscillator = 1/2 mv^2 ?so we assume KE of single linear oscillator = PE of it. so we have 2 times?

so total energy of circular motion = total energy of oscillator?

This is not quite true. The total energy is KE +PE, but when one of them is maximum, the other is zero.
For a body preforming SHM, x=Asin (wt). The potential energy is maximum at x=A, and PE(max) = 1/2kA^2. The speed is maximum at x=0, and v(max)=Aw, the maximum kinetic energy = 1/2 m A^2 w^2. As w^2=k/m, KE(max)=1/2 kA^2, the same as the maximum potential energy.

Two orthogonal vibrations with the same frequency and amplitude and phase difference of pi/2 between them is equivalent with a circular motion. Because of the phase difference, when one oscillator has maximum potential energy, the other has its maximum kinetic energy, but they are equal, so the energy of this circular motion is equal to that of two oscillators.

Note that this is true only for such circular motions which result from these two vibrations. For the planetary motion, or motion in Coulomb field, both the force and the energy is different.

ehild

 

[quietrain]

oh i see! so when i use mvr = nh/2pi , i am acutally accounting for both the KEmax and the PEmax of the circular motion? that's why i have to divide by 2 if i want to find just one oscillator?

 

[ehild]

Something like that :). I think you haven't studied quantum theory yet, have you? Because the derivation of the energy of quantum harmonic oscillator is at advanced level. Perhaps the formula was shown already during the classes.

ehild

 

[quietrain]

yup thanks a lot!