Answer to Question #249428 in Physics for Afroja Akhi

Question

An underdamped harmonic oscillator has its amplitude reduced to

th of its initial
value after 100 oscillations. If the time period be 1.15 sec, calculate the time in which
(i) its amplitude, and (ii) energy falls to

th of its undamped value. If the mass of the
oscillator be 1.127 gm, calculate its average rate of loss of energy

Answer 1

Given:

$A=\frac{1}{10}A_0$

$N=100$

$T=1.15 \:\rm s$

$m=1.127\:\rm g$

(i) The law of oscillations

$$A=A_0e^{-kt}$$

After 100 oscillations

$$A=\frac{1}{10}A_0=A_0e^{-100kT}$$

Then

$$k=\frac{\ln 10}{100T}=\frac{\ln 10}{100*1.15}=0.02\:\rm s^{-1}$$

So, for $A=A_0/e$

$$A_0/e=A_0e^{-0.02t}$$

Corresponding time

$$t=\frac{\ln e}{0.02}=50\:\rm s$$

(ii) The energy is proportional to the amplitude suared, so

$$E_0/e=E_0e^{-2*0.02t}$$

Corresponding time

$$t=\frac{\ln e}{2*0.02}=25\:\rm s$$

The energy loss rate

$$\frac{dE}{dt}=-2kE_0e^{-2kt}$$