Answer
$21\;\rm oscillation$
Work Step by Step
Our system undergoes a damping simple harmonic motion, so the maximum displacement is given by
$$A_{(t)}=Ae^{−t/2\tau}$$
Hence, we need to find the time at which the final amplitude is $e^{-1} $ the initial amplitude.
$$e^{-1}\color{red}{\bf\not} A =\color{red}{\bf\not} Ae^{−t/2\tau}$$
Thus,
$$\dfrac{-t}{2\tau}=-1$$
$$t=2\tau=\dfrac{2m}{b}$$
$$t=\dfrac{2m}{b}\tag 1$$
Now we need to find the period of one full oscillation to find the number of oscillations after time interval of $t=33.33\;\rm s$.
We know that the angular frequency of a damped oscillator is given by
$$\omega=2\pi f=\sqrt{\dfrac{k}{m}-\dfrac{b^2}{4m^2}}$$
where $f=1/T$, so
$$ \dfrac{2\pi}{T}=\sqrt{\dfrac{k}{m}-\dfrac{b^2}{4m^2}} $$
So,
$$ T= \dfrac{2\pi}{\sqrt{\dfrac{k}{m}-\dfrac{b^2}{4m^2}} } \tag 2 $$
The number of oscillations is given by
$$N=\dfrac{t}{T}$$
Plugging from (1) and (2);
$$N=\dfrac{2m/b}{\dfrac{2\pi}{\sqrt{\dfrac{k}{m}-\dfrac{b^2}{4m^2}} } }$$
$$N=\dfrac{2m\sqrt{\dfrac{k}{m}-\dfrac{b^2}{4m^2}} }{ 2b\pi }$$
$$N=\dfrac{2m}{2b\pi }\sqrt{\dfrac{k}{m}-\dfrac{b^2}{4m^2}} $$
Plugging the known;
$$N=\dfrac{2(0.250)}{2(0.015)\pi }\sqrt{\dfrac{4}{0.250}-\dfrac{(0.015)^2}{4(0.250)^2}}=\bf 21.22 $$
$$N\approx\color{red}{\bf 21}\;\rm oscillation$$
So it needs 21 oscillations to reach the amplitude to $e^{-1}A$.
