Quantum Mechanics pendulum problem

[bon]

Homework Statement

The pendulum of a grandfather clock has a period of 1s and makes excursions of 3cm either side of dead centre. Given that the bob weighs 0.2kg, around what value of n would you expect its non negligible quantum amplitudes to cluster?

Homework Equations

The Attempt at a Solution

I think the n here refers to the nth energy eigenvalue so En = (n + 1/2) h/2pi w

How do i work this out? My guess is that I need to work out the energy of a classical harmonic oscillator and equate this to (n+1/2) h/2pi w to get n?

So i know w = 2pi ... but how do i work out the energy of the oscillator? At max displacement it will have no KE, only PE..but how do i work out what this is?

Thanks!

 

[graphene]

dude, draw a picture with both the equilibrium and final positions, then find out how high the final position is w.r.t to the equilibrium position.

 

[bon]

dude, draw a picture with both the equilibrium and final positions, then find out how high the final position is w.r.t to the equilibrium position.

thanks..but this is my problem. i don't know how high the final position is wrt initial position... all they say is that it makes excursions of 3cm either side of dead centre...but i don't know what what angle it is at at the extrema, so how can i work out how high it goes?

Thanks

 

[___]

If you have the period, don't you have the length? T = 2pi sqrt(l/g) approximately.
From this you can get the height? I have no knowledge of quantum mechanics so this is just a shot in the dark.

 

[bon]

Thanks, but l in that equation gives you the length of the string. I still don't see how you can get the height..

 

[___]

Thanks, but l in that equation gives you the length of the string. I still don't see how you can get the height..

Again, just a shot in the dark, but by height I'm guessing you mean the height the mass reaches at it's full swing?
You know its 3 cm far, and l long. The height is x= 3 /(tan ( 90 - arctan(3/l))

 

[bon]

ok thanks so i get h = 1.79 x 10^-3m

is this right?

This means E total = mgh = 0.2 * 9.8 * h = 3.5 x 10^-3 J

so now do i set (n+1/2) h/2pi w = 3.5 x 10^-3 ?

This gives n as something ridiculously large (5.29 x 10 ^30). Is this right?

 

[___]

The equation I gave was wrong, sorry man. I have a new relation:

9 - 2Lx + x^2 = 0

which gives

x = L +- sqrt(4L^2 - 36) / 2This comes from considering it's equilibrium and max and getting an isosceles triangle.
You split it into 2 right angle triangles. Use pythagoras to get 3^2 + (L-x)^2 = L^2

 

[bon]

Oh ok - i worked it out myself, using a different method, but i think the answer is the same, right?

Is my final answer correct then (to someone who knows about the quantum side of the question as well...)?

 

[bon]

anyone?

 

[bon]

it just seems the number n is too big... have i interpreted the question correctly?