Damped oscillator with 5% energy loss per cycle

[e(ho0n3]

Question: A damped harmonic oscillator loses 5.0 percent of its mechanical energy per cycle. (a) By what percentage does its frequency differ from the natural frequency $\omega_0 = \sqrt{k/m}$? (b) After how may periods will the amplitude have decreased to 1/e of its original value?

So, for (a), the answer is $\omega ' / \omega_0$ where

$$\omega ' = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2}}$$

But that leaves me with 3 unknowns, k, m, and b requiring three equations to solve. The only equations I can think of is E = K + U (mechanical energy) and E = 0.95^TE₀ where T is the number of cycles and E₀ is the initial mechanical energy.

What other equation can I use? Or is there a simpler method of finding the solution?

 

[Andrew Mason]

Question: A damped harmonic oscillator loses 5.0 percent of its mechanical energy per cycle. (a) By what percentage does its frequency differ from the natural frequency $\omega_0 = \sqrt{k/m}$? (b) After how may periods will the amplitude have decreased to 1/e of its original value?

So, for (a), the answer is $\omega ' / \omega_0$ where

$$\omega ' = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2}}$$

But that leaves me with 3 unknowns, k, m, and b requiring three equations to solve. The only equations I can think of is E = K + U (mechanical energy) and E = 0.95^TE₀ where T is the number of cycles and E₀ is the initial mechanical energy.

What other equation can I use? Or is there a simpler method of finding the solution?

The amplitude is given by:

$$A = A_0e^{-\gamma t}$$ where $\gamma = b/2m$ ¹

Work out the value for $\gamma$ given that A^2 decreases to .95A_0^2 in the first $T = 2\pi /\omega '$ seconds.

Then find $\omega '$ in terms of $\omega_0$ using:
$\omega '^2 = \omega_0^2 - \gamma^2$

AM

[Note: 1. The solution to the damped harmonic oscillator is:

$$x = A_0e^{-\gamma t}sin(\omega 't + \phi)$$

where $\omega ' = \sqrt{\omega^2 - \gamma^2}$ ]

 

[e(ho0n3]

Work out the value for $\gamma$ given that A^2 decreases to .95A_0^2 in the first $T = 2\pi /\omega '$ seconds.

How do you know the square of the amplitude decreases to .95A_0^2 in the first T seconds?

 

[Andrew Mason]

How do you know the square of the amplitude decreases to .95A_0^2 in the first T seconds?

Energy is proportional to the square of the amplitude (all energy is potential energy at maximum amplitude: $E = \frac{1}{2}kx^2$). If the system loses 5% of its energy in one cycle, the square of the amplitude will decrease to 95% of the square of the original amplitude.

AM