A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to $$\frac{1}{1000}$$ of the original amplitude is close to:

We have a damped harmonic oscillator whose amplitude decreases exponentially with time. The mathematical form is

$$A(t)=A_0\,e^{-\beta t},$$

where $$A_0$$ is the initial amplitude and $$\beta$$ (read “beta”) is the damping constant. Our task is to determine how much time the system needs for the amplitude to fall to $$\dfrac{1}{1000}$$ of its initial value.

First, let us translate the information given into mathematical equations. The frequency of oscillation is 5 Hz, which means the system completes 5 oscillations every second. Equivalently, the time period of one oscillation is

$$T=\dfrac{1}{f}=\dfrac{1}{5}\ \text{s}=0.2\ \text{s}.$$

We are told that the amplitude becomes half of its value after every 10 oscillations. Ten oscillations take a time interval of

$$t_1=10T=10\times 0.2\ \text{s}=2\ \text{s}.$$

At this instant the amplitude is

$$A(t_1)=\dfrac{A_0}{2}.$$

Substituting $$t=t_1$$ in the exponential decay formula gives

$$A(t_1)=A_0\,e^{-\beta t_1}=\dfrac{A_0}{2}.$$

Cancelling $$A_0$$ from both sides, we obtain

$$e^{-\beta t_1}=\dfrac{1}{2}.$$

Taking the natural logarithm (base e) of both sides, we get

$$-\beta t_1=\ln\!\left(\dfrac{1}{2}\right)=-\ln 2.$$

Therefore, the damping constant is

$$\beta=\dfrac{\ln 2}{t_1}=\dfrac{\ln 2}{2}.$$

Now we need the time $$t_2$$ at which

$$A(t_2)=\dfrac{A_0}{1000}.$$

Using the decay law again,

$$A(t_2)=A_0\,e^{-\beta t_2}=\dfrac{A_0}{1000}.$$

Dividing by $$A_0$$ and taking natural logarithms:

$$e^{-\beta t_2}=\dfrac{1}{1000}\quad\Longrightarrow\quad -\beta t_2=\ln\!\left(\dfrac{1}{1000}\right)=-\ln 1000.$$

Thus

$$t_2=\dfrac{\ln 1000}{\beta}.$$

We already have $$\beta=\dfrac{\ln 2}{2},$$ so substituting this value gives

$$t_2=\dfrac{\ln 1000}{\dfrac{\ln 2}{2}}=\dfrac{2\ln 1000}{\ln 2}.$$

Now we compute the logarithms. Using the facts $$\ln 1000=\ln 10^3=3\ln 10$$ and $$\ln 10\approx 2.302585,$$ we get

$$\ln 1000 = 3\times 2.302585 \approx 6.907755.$$

Also, $$\ln 2\approx 0.693147.$$ Therefore,

$$t_2 \approx \dfrac{2\times 6.907755}{0.693147} =\dfrac{13.81551}{0.693147} \approx 19.93\ \text{s}.$$

This value is very close to 20 s, matching the option provided. Hence, the correct answer is Option C.