# Frequency of a tuning fork in a vacuum

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.

## 2 Answers

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

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

• 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

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.