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I've been having trouble with my physics homework. The problem is:

You may have measured the properties of a simple spring-mass system in the lab. Suppose you found ks = 0.9 N/m and m = 0.01 kg, and you observed an oscillation with an amplitude of 0.5 m. What is the approximate value of N, the "quantum number" for this oscillator? (That is, how many levels above the ground state is this oscillator?)

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closed as too localized by David Z Oct 29 '12 at 5:22

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1 Answer 1

First solve the differential equation

$$ \frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 2\zeta\omega_0\frac{\mathrm{d}x}{\mathrm{d}t} + \omega_0^2 x = \frac{1}{m} F_0 \sin(\omega t),$$

You then get

$$x(t)=A \cos(t \sqrt{\frac{k}{m}})$$

Then solve for x. You know the potential energy is $V=\frac{1}{2} k x^{2}$. Calculate this value then set it equal to $V= \hbar \omega(\frac{1}{2} +n)$. Then you can estimate the level of this oscillator. Take t for one or something.

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It should be a big number (10^xx), which should be obvious I think given that this is well within the classical limit. –  Dylan Sabulsky Oct 29 '12 at 4:54
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