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**(a)**Show that the total mechanical energy of a lightly damped harmonic oscillator is

[tex]E = E_0 e^{-bt/m}[/tex]

where [itex]E_0[/itex] is the total mechanical energy at t = 0.

**(b)**Show that the fractional energy lost per period is

[tex]\frac{\Delta E}{E} = \frac{2 \pi b}{m \omega_0} = \frac{2 \pi}{Q}[/tex]

where [itex]\omega_0 = \sqrt{k/m}[/itex] and [itex]Q = m \omega_0 / b[/itex] is called the quality factor or Q value of the system. A larger Q value means the system can undergo oscillations for a longer time.

My Answer:

**(a)**When the velocity of the oscillator is 0, the total mechanical energy is purely potential energy, [itex]U = 1/2kx^2[/itex]. Since I know that [itex]x = Ae^{-bt/(2m)}\cos{\omega't}[/itex] where t is some multiple of [itex]2\pi/\omega'[/itex], then

[tex]E = \frac{1}{2}kA^2e^{-bt/m}[/tex]

and [itex]E_0 = 1/2kA^2[/itex]. Of course, this is only valid when the velocity of the oscillator is 0, but since it is lightly damped the total mechanical energy should be approximately the same when the velocity is > 0. Right?

**(b)**Using some calculus, I can timidly state that

[tex]\frac{\Delta E}{\Delta t} \approx \frac{dE}{dt} = -\frac{E_0b}{m}e^{bt/m}[/tex]

and since [itex]\Delta t = 2\pi / \omega'[/itex] then

[tex]\frac{\Delta E}{E} = -\frac{b\Delta t}{m} = -\frac{2\pi b}{m\omega'}[/tex]

Since the oscillator is lightly damped, [itex]\omega' \approx \omega_0[/itex]. However the result I get is negative. Should it be negative?