Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Consider this equation of a damped harmonic oscillator such that: $$ \ddot{x}+2\gamma\dot{x}+\omega^2_0=0 $$

with: $\gamma=\frac{b}{2m}$ and $\omega_0=\sqrt{\frac{k}{m}}$

Finally, we know that the equation x(t) should be of this form: $$ x(t)=e^{-\gamma t}[Acos(\omega_1t)+Bsin(\omega_1t)] $$

It is observed that the amplitude of oscillation of a tuning fork of frequency 400Hz is damped in the air by 10% in 12s. What would be the frequency of the tuning fork in a vacuum (void)?

I'm really struggling to find a starting point… I've found that: $$ \omega_1=\frac{\sqrt{4mk-b^2}}{2m} $$ But I don't see where to begin to find the frequency in the vacuum. Could someone explain me how to start?

Thank you in advance.

share|improve this question
1  
en.wikipedia.org/wiki/… tells you the relation between the damped and undamped frequencies. –  John Rennie Oct 17 '12 at 17:14
add comment

2 Answers

up vote 0 down vote accepted

From the amplitude decrement you have $0.9 = {\rm e}^{-\gamma\, 12}$ or $\gamma = 0.00878$

From the damped frequency you have $\omega_1 = 2 \pi\; 400 = 2513.27 \,{\rm rad/s} $

You have already stated that $\gamma = \frac{b}{2 m}$ as well as

$$ \omega_1^2 = \frac{k}{m} - \frac{b^2}{4 m^2} $$ $$ \omega_1^2 = \frac{k}{m} - \gamma^2 $$

Note that the undamped oscillation is $\omega_0 = \sqrt{\frac{k}{m}} $ so

$$ \omega_1^2 = \omega_0^2 - \gamma^2 $$ $$ \omega_0 = 2513.27 {\rm rad/s} = 399.999 {\rm Hz} $$

share|improve this answer
    
Note that the above gives a damping ratio of $3.5\; 10^{-6}$ which is to say zero. –  ja72 Oct 17 '12 at 17:41
    
Thank you, it looks really easy now! –  Oliver Oct 17 '12 at 17:57
add comment

Forget about the relation to mass and spring constant, $$ \omega_0 = \sqrt{\omega_1^2-\gamma^2} $$ is all you need. $\omega_1=2\pi f$ and the damping rate if found from the observed decay.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.