A $200g$oscillator in a vacuum chamber has a frequency of $2.0Hz$. When air is admitted, the oscillation decreases to $60\%$of its initial amplitude in $50s$. How many oscillations will have been completed when the amplitude is $30\%$of its initial value?

Damped Oscillations: During oscillations in a true oscillating system, mechanical energy $E$decreases because external forces, such as drag, inhibit oscillations and convert mechanical energy to thermal energy. The real oscillator, as well as its motion, is then said to be damped. The oscillator's displacement is provided by if the damping force is given by $F=-bv$, where is the oscillator's velocity and $b$is the damping constant.

$x\left(t\right)=A{e}^{-bt/2m}\cos (\omega t+\varphi )$

where $\omega$denotes the angular frequency of the damped oscillator, is given by

$\omega =\sqrt{\frac{k}{m}-\frac{b^2}{4m^2}}$

The angular frequency of an undamped oscillator $(b=0)$is .

A basic harmonic oscillator's period $T$is proportional to its oscillation frequency $f$as follows:

$T=\frac{1}{f}$

Given data:

The oscillator has a mass of:$m=(200\mathrm{g})\left(\frac{1\mathrm{kg}}{1000\mathrm{g}}\right)=0.200\mathrm{kg}$

The oscillator's frequency of oscillation is:$f=2.0\mathrm{Hz}$

The oscillator's amplitude reduces to $60\%$of its starting value over a time interval of $t=50s$.

When the oscillator's amplitude drops to $30\%$of its starting value, we must calculate the number of oscillations completed.

solution:

As a function of time, the amplitude of the ball's oscillation is expressed as follows:

$A\left(t\right)=A_0{e}^{-bt/2m}$

where $A_0$is the oscillator's initial amplitude. Rearrange the following and solve for $b$:

$\frac{A\left(t\right)}{A_0}=e^{-bt/2m}$

$\ln \left[\frac{A\left(t\right)}{A_0}\right]=-\frac{bt}{2m}$

$b=-\left(\frac{2m}{t}\right)\ln \left[\frac{A\left(t\right)}{A_0}\right]$

We substitute $60\%$$A_0$for $A\left(t\right)$because the oscillator's amplitude decreases to $60\%$of its starting value $A_0$at time $t=50s$:

$b=-\left(\frac{2m}{t}\right)\ln \left(\frac{60\%A_0}{A_0}\right)$

$=-\left(\frac{2m}{t}\right)\ln (60\%)$

Substitute the following numerical values:

$b=-\left[\frac{2(0.200\mathrm{kg})}{50\mathrm{s}}\right]\ln (60\%)$

$=4.0866\times {10}^{-3}\mathrm{kg}/\mathrm{s}$

Solving Equation $\left(3\right)$for $t$yields the time it takes for the oscillator's amplitude to drop to $30\%$of its starting value:

$\frac{A\left(t\right)}{A_0}=e^{-bt/2m}$

$\ln \left[\frac{A\left(t\right)}{A_0}\right]=-\frac{bt}{2m}$

$t=-\frac{2m}{b}\ln \left[\frac{A\left(t\right)}{A_0}\right]$

We substitute $30\%A_0$for $A\left(t\right)$because we are looking for the period when the amplitude reduces to $30\%$of its original values:

$t=-\frac{2m}{b}\ln \left(\frac{30\%A_0}{A_0}\right)$

$=-\frac{2m}{b}\ln (30\%)$

Substitute the following numerical values:

$t=-\frac{2(0.200\mathrm{kg})}{4.0866\times {10}^{-3}\mathrm{kg}/\mathrm{s}}\ln (30\%)$

$=117.8457724\mathrm{s}$

Equation $\left(2\right)$is used to calculate the oscillator's period of oscillation:

$T=\frac{1}{f}$

Substitute the following numerical values:

localid="1650095449986" $T=\frac{1}{2.0\mathrm{Hz}}$

localid="1650095454101" $=0.5\mathrm{s}$

The oscillator's total number of oscillations in time $t$is then calculated.

$n=\frac{t}{T}$

$=\frac{117.8457724\mathrm{s}}{0.5\mathrm{s}}$

$\approx 236$oscillations