Quality of a Tuning Fork

If the sound intensity decreases by a factor of 4 in 6 seconds, then:

$e^{-6b/m} = \frac14.$

where $b$ and $m$ are the damping constant and mass of the tuning fork respectively treating it as a damped harmonic oscillator .

Taking the logarithm of both sides and rearranging,

$\frac{m}{b} = \frac{6}{\log 4}.$

By the definition of the quality factor:

$Q = \frac{m\omega}{b},$

where $\omega$ is measured in radians. Since $\omega = 261.6 \text{ Hz}$ , find:

$Q = \frac{6}{\log 4} (261.6)(2\pi) \approx 7114.$