# Swinging Pendulum (Quantum)

You are given a swinging pendulum that has a length of $l=10 cm$. This pendulum oscillates with a small amplitude. The amplitude of this swinging pendulum is $A=1 cm$ (when $\theta=0.1 rad$) for a mass of $m=1 gm$

a) Please find the total energy in ergs of this pendulum.

b) Please find the minimum energy quanta of this pendulum (in ergs).

c) For one quanta absorbed or released please find the change in amplitude of this pendulum of $l=10 cm$.

d) Please find the number of energy levels of this pendulum when the amplitude changes from A=0 to A=1 cm.

e) If $h= 1 erg-sec$ instead of $h=6.6*10^{-27} erg-sec$ please find the answers to b, c, and d.

#### Swinging Pendulum Answer

For small oscillations we write

where we have made the approximation of: $\sin(\theta) = \tan(\theta) = \theta$

The period and frequency of this pendulum is therefore

$T = 2*\pi*\sqrt{L/g} = 2*\pi*\sqrt{0.1/9.80} =0.6347 s$

$\omega = 2*\pi/T = 9.899 rad/s$

a)

The total energy of the pendulum is constant and is purely kinetic at the bottom of the trajectory:

$E = m*v^2/2 =(1/2)*m*(\omega*L)^2 = (1/2)*10^{-3}* (9.899*0.1)^2 = 4.9*10^{-4} J =4900 erg$

($1 erg =10^{-7} J$)

b)

One quanta of energy is thus

$E0 = h*F = h/T = 6.62*10^{-34}/0.6347 =1.04*10^{-33} J= 1.04*10^{-26} ergs$

c)

by differentiation of the expression of energy in a) we obtain

$dE = m*\omega^2*L*dL$

and thus the difference in the amplitude of oscillation is

$dL = dE/(m*\omega*L) =1.04*10^{-33}/(10^{-3}*9.899*0.1) = 1.05*10^{-30} m$

d)

The total number of energy levels between $A =0$ and $A =1 cm$ is

$N = A/dL = 0.01/1.05*10^{-30} =9.5*10^{27} levels$

In the case when the Planck constant has the value of $h =1 erg*s=10^{-7} J*s$ we have

$E0 =1.576*10^{-7} J = 1.576 erg$

$dL =1.59*10^{-4} m$

$N =62.8 levels$

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