# SHM Oscillation problem involving potential energy

## Homework Statement

A molecular bond can be modeled as a spring between two atoms that vibrate with simple harmonic motion. Figure P14.63 shows an SHM approximation for the potential energy of an HCl molecule. For E < 4 * 10^-19 J it is a good approximation to the more accurate HCl potential-energy curve that was shown in Figure 10.31. Because the chlorine atom is so much more massive than the hydrogen atom, it is reasonable to assume that the hydrogen atom (m = 1.67 * 10^-27 kg) vibrates back and forth while the chlorine atom remains at rest. Use the graph to estimate the vibrational frequency of the HCl molecule.

## Homework Equations

ƒ = (1/2PI) * sqrt(k/m)
Umax = (1/2)*k*A^2

## The Attempt at a Solution

I figured from the graph that the Amplitude (max. displacement) equals 0.17nm - 0.13 nm = 0.04 nm, which equals 4 x 10^-11 meters.

It also appears from the graph that the max. potential energy is 4 x 10^-19 J. So:

Umax = (1/2)*k*A^2
4x10^-19 J = (1/2)*k*A^2
k = (4x10^-19 J) * 2 / A^2 = (4x10^-19 J) * 2 / (4x10^-11 m)^2 = 500 N/m

So, frequency = (1/2∏) * sqrt(k/m) = (1/2∏) * sqrt(500 N/m / 1.67x10^-27 kg) = 8.71x10^13 Hz

But the answer key says 7.9x10^13 Hz. I'm not sure where I went wrong. I would much appreciate any help.

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Simon Bridge
Homework Helper
OK - so the question is:
Use the graph to estimate the vibrational frequency of the HCl molecule.
You have pulled the amplitude ##A## and max PE ##U_{max}## off the graph, then used the formula ##U_{max}=\frac{1}{2}kA^2## to compute ##k##.

You got the right order of magnitude - so it may just be a matter of the choice of value for something.
i.e. check your value for the amplitude more carefully - I get a different number to you using a set-square to line up the axis.

Note: it is best practice to do all the algebra before you put numbers in.
i.e. derive the equation that related the frequency with what you know.

OK - so the question is:
You have pulled the amplitude ##A## and max PE ##U_{max}## off the graph, then used the formula ##U_{max}=\frac{1}{2}kA^2## to compute ##k##.

You got the right order of magnitude - so it may just be a matter of the choice of value for something.
i.e. check your value for the amplitude more carefully - I get a different number to you using a set-square to line up the axis.

Note: it is best practice to do all the algebra before you put numbers in.
i.e. derive the equation that related the frequency with what you know.

OK, I took another look at the graph and it seems like the amplitude might actually be closer to 0.045 nm = 4.5 x 10^-11 m...

So using the same equation as before, I get:

k = 2/A^2 * Umax = 2/(4.5x10^-11 m)^2 * 4x10^-19J = 395.06 N/m.

ƒ = 1/(2∏) * √(k/m)
= 1/(2∏) * √( (395.06 N/m) / (1.67x10^-27 kg) )
= 7.74x10^13 Hz

So my answer now is closer to the one in the answer key (7.9x10^13 Hz) than before, but still off by a bit. Aside from the possible mistake in measuring the amplitude from the graph, is there any other possible (more fundamental) mistake I might've made anywhere?

Just as an aside, I figured there's an alternative way to get the answer...
There's a formula that says Vmax = A*ω. We can begin by getting the value of Vmax via the following energy equation:

-> (1/2) * m * Vmax^2 = Umax
-> Vmax = √(2 * Umax / m) = √(2 * 4x10^-19J / 1.67x10^-27 kg) = 21887 m/s

So we solve for ω:
-> Vmax = A*ω
-> ω = Vmax / A = (21887 m/s) / (4.5x10^-11 m) = 4.8638 rad/s
-> Convert ω to ƒ -> (4.8638 rad/s) / (2∏) = 7.74x10^13 Hz

Again I get the same answer as before...have I made any mistakes in any of my computations, aside from possibly plugging in the wrong numbers? Thanks.

Simon Bridge
Homework Helper
Any of you chosen values could be out.
Double check - why did you choose that particular value?
Think it through - which way to the values need to change to make a difference?

There is always more than one way to do things - best practice is still to do all the algebra first, then plug in the numbers. You are still refusing to do that for some reason. One of the sources of small mistakes is in implicit rounding off.

The "new" approach gets you to the same exact equation as before - there is no difference.

Finally - you have done just what I'd do.
It is always possible that the model answer is wrong ;)