Answer to Question #139931 in Physics for Jisha

Question

4. The quality factor Q of a tuning fork is 5.0 x 104. Calculate the time interval after which its energy becomes (1/10)th of its initial value. The frequency of the tuning fork is 300 sec-1. (loge 10=2.3)

Answer 1

The energy magnitude is proportional to:

$$E \propto \exp\left(-\frac{1}{Q}\omega t\right)$$

where $\omega = 2\pi\times 300s^{-1} = 600\pi$

Let $t_1$ be the initial moment of time and $t_2$ be the final one. Thus, the ratio of energies (which is eventually equal to 1/10) is:

$$\dfrac{E_{final}}{E_{initial}} = \dfrac{ \exp\left(-\frac{1}{Q}\omega t_2\right)}{ \exp\left(-\frac{1}{Q}\omega t_1\right)} = \exp\left(-\frac{1}{Q}\omega (t_2-t_1)\right) = \\ = \exp\left(-\frac{1}{Q}\omega \Delta t\right) = \dfrac{1}{10}$$

where $\Delta t$ is an required time interval. Expressing $\Delta t$ from the last equation, get:

$$\Delta t = \dfrac{Q\ln10}{\omega} = \dfrac{5\times 10^4\cdot 2.3}{600\pi} \approx 602.14s$$

Answer. 602.14 s.