Answer to Question #89779 in Electricity and Magnetism for Shivam Nishad

Question

The quality factor of a weakly damped
oscillator of frequency 200 Hz is 1000.
Calculate the time in which its energy
becomes 20% of the initial value.

Answer 1

Given:

$$Q = 1000$$

$$f = 200 Hz$$

$$\frac{E_t}{E_0} = 20\% = 0.2$$

Solution:

The quality factor for weakly damped oscillator is :

$$Q = \frac{\omega _0}{\gamma}$$

So, damping rate is:

$$\gamma = \frac{\omega _ 0 }{Q}$$

Energy of such oscillator is:

$$E(t) = E_0 e^{-\gamma t}$$

Solving for t:

$$E_t = E_0 e^{-\frac{\omega _0 t}{Q}}$$

$$\frac{E_t}{E_0} = e^{-\frac{\omega _0 t}{Q}}$$

$$ln(\frac{E_t}{E_0} )= -\frac{\omega _0 t}{Q}$$

Finally:

$$t = -\frac{Q}{\omega _0} ln(\frac{E_t}{E_0}) = -\frac{Q}{2 \pi f} ln(\frac{E_t}{E_0})$$

Calculating:

$$t \approx -\frac{1000}{2 * 3.14 * 200} ln(0.2) \approx 1.6 *\frac{1000}{1256} = 1.27 sec$$

Answer: 1.27 sec