Small Oscillations: Homework on Atom Mass, Earth Gravity

[CaptainEvil]

Homework Statement

Consider an atom of mass m bonded to the surface of a much larger immobile body by
electromagnetic forces. The force binding the atom to the surface has the expression

F = e^{acosz + bsinz} + dtanz

where a, b and d are constants and z is positive upwards. The equilibrium point is defined
to be the origin, so z=0 there. Ignore any motion except in the vertical direction. The
whole assembly is subject to normal Earth gravity.

a) For small oscillations, give an approximate expression for the binding force on the
atom.
b) What restrictions are there on the values of a, b and d so that the force on the atom
actually is a restoring force and the atom can reach a stationary equilibrium?
c) What is the angular frequency ω0 and frequency ν0 of oscillation of this undamped
system?
d) Would the oscillation frequency change if there were no gravity? Why?
e) Subject to the considerations above, if the atom has a mass of 1 atomic mass unit,
30 b = −1.658×10 and 4 d = −3.2361×10 N, what is the frequency ν0 of the oscillation?
f) Suppose that a photon with this frequency that is incident on this atom would be
absorbed. What wavelength does this correspond to? What part of the spectrum does it
fall in?

Homework Equations

The Attempt at a Solution

I have completed the problem, and think I'm correct. Any confirmation, thoughts would be greatly appreciated!

a) in the z component, we can simplify the expression as a taylor series, disregarding the z terms with higher power terms than 2.

F(z) = F₀ + (dF/dz)₀z

F₀ = 0 at equilibrium so F(z) is simply the first derivative of F evaluated at 0

=(be^a + d)z

b) restoring force means that k (spring constant) must be negative.

so be^a + d < 0

re-arranging yields that d < 0, b > 0 and a < ln(-d/b)

c) w<sub>0/SUB] = $$\sqrt{k/m}$$ where k = be^a + d

and v0 = w0/2$$\pi$$

d) No, because with microscopic distances and particles, electromagnetic force dominates over gravity, so gravity has a negligible effect.

e) Need to solve for a, so under equilibrium, force must balance gravity so F0 = mg. Fo from original eqn is e^a so we can solve for a.

a = ln(mg) = 64.96 N

plugging in then yields vo as 6.42 x 10¹⁴ Hz, which is 467.3 nm, in the visible.

Thoughts?</sub>

 

[gabbagabbahey]

a) in the z component, we can simplify the expression as a taylor series, disregarding the z terms with higher power terms than 2.

F(z) = F₀ + (dF/dz)₀z

F₀ = 0 at equilibrium so F(z) is simply the first derivative of F evaluated at 0

=(be^a + d)z

Careful, equilibrium means that the net force on the particle is zero...there is more than just the binding force acting on the particle...what do you really get when you plug the equilibrium point into your original expression for the binding force?

 

[CaptainEvil]

I have taken that into account, in part e. under equilibrium, binding force is balancing with gravity. I am quite sure part a is correct.

 

[gabbagabbahey]

I'm quite sure part (a) is incorrect...the question asks for an approximate expression for the binding force, not the net force...$F_{net}(z=0)=0$, but $F_{binding}(z=0)=$____?

 

[CaptainEvil]

my textbook (that the prof follows very closely) states that for small oscillations it is acceptable to expand as a taylor series, disregard the powers of z higher than 1, and that Fo vanishes from equilibrium, and you're left with the first derivative.

If I'm wrong, please explain.

 

[gabbagabbahey]

The problem is that you are asked to expand the binding force for small equilibriums...F₀ only equals zero, if F₀ is the net force at equilibrium...the binding force does not need to be zero at equilibrium.

Plug z=0 into you original expression for the binding force...and into your answer for part (a)...do you get the same result?

 

[CaptainEvil]

so you're saying F(z) = aexpe + bexpa + d ? the aexpe term being binding force at 0.

 

[CaptainEvil]

sorry that's expa not aexpe

 

[gabbagabbahey]

so you're saying F(z) = aexpe + bexpa + d ? the aexpe term being binding force at 0.

Close, Taylor expanding the binding force around the equilibrium point ($z=0$) gives you:

$$F_{binding}(z)\approx F_{binding}(0)+F'_{binding}(0)z=e^a+(be^a+d)z$$

...make sense?

Now redo part (b).

 

[CaptainEvil]

doesn't b) remain the same? because only k needs to be negative yes? which is just the part attatched to the z?

 

[gabbagabbahey]

b) What restrictions are there on the values of a, b and d so that the force on the atom
actually is a restoring force and the atom can reach a stationary equilibrium?

Don't forget about the last part of this question...what is the net force? What must it be for the particle to be in equilibrium at $z=0$? That should tell you the value of $a$.

b) restoring force means that k (spring constant) must be negative.

so be^a + d < 0

Right.

re-arranging yields that d < 0, b > 0 and a < ln(-d/b)

No, this doesn't follow from your previous statement any more than $2<0$, $-2>0$ and $0<\ln(-2/-2)$ would follow from the equation $-2e^0+2<0$

 

[CaptainEvil]

Right, I solved for a in part (e), so I bring that down, and say that to reach an equilibrium, gravity must balance the electric force. Then I can solve for a.

Fo = exp(a) = mg. so
a = ln(mg) = 64.96 N

As for restrictions on b and d, is it not correct that the ratio of -d/b needs to be positive?

 

[gabbagabbahey]

Right, I solved for a in part (e), so I bring that down, and say that to reach an equilibrium, gravity must balance the electric force. Then I can solve for a.

Fo = exp(a) = mg. so
a = ln(mg) = 64.96 N

Good, but I wouldn't use the numeric value of 64.96N until part (e)...just leave it as a=ln(mg) for now...

As for restrictions on b and d, is it not correct that the ratio of -d/b needs to be positive?

Yes, but that doesn't mean it has to be greater than 'a'...plugging in a=ln(mg), you get $mgb+d<0$...that's all you can really say about 'b' and 'd'.

 

[CaptainEvil]

great thanks! do all the other answers look okay?

 

[gabbagabbahey]

You should definitely rethink part (d)...what is the net force when there is no gravity...will that net force actually result in oscillations at all?

 

[CaptainEvil]

ah yes, I see now that because a is dependent on g, and a is in the expression for oscillation frequency, that it will be affected if gravity should be neglected.

Is this correct?

 

[gabbagabbahey]

Well yes, but how will it be affected?