Question

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 $${\displaystyle \frac{1}{1000}}$$ of the original amplitude is close to:-

A. 100 s
B. 20 s
C. 10 s
D. 50 s

Answer 1

Hint: As, the time taken by the oscillator is given to be dropped to half the initial value, thus, to solve this problem, we will make use of the formula that relates the half-life with a decay constant in terms of the amplitude of the oscillation and time taken.
Formula used:
$$A=A_0{e}^{-\gamma t}$$

Complete answer:
The formula that relates the amplitude of the oscillations produced by an oscillator with the time taken by the oscillator is given as follows.
$$A=A_0{e}^{-\gamma t}$$
Where A is the amplitude of the oscillation, t is the time taken, $$\gamma$$ is the decay constant and $$A_0$$ is the amplitude after a drop of the original amplitude.
From given, we have the data,
The frequency of the damped harmonic oscillator is 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations.
Substitute these values in the above equation. So, we have,
$$\begin{array}{rl}& A=A_0{e}^{-\gamma t}\\ & \Rightarrow {\displaystyle \frac{A_0}{2}}=A_0{e}^{-\gamma (2)}\end{array}$$
Continue the further calculation, and express the equation in terms of the decay constant.
$$\begin{array}{rl}& 2=e^{2\gamma }\\ & \Rightarrow \gamma ={\displaystyle \frac{\ln 2}{2}}\end{array}$$
Now let us compute the time taken by the damped harmonic oscillator to drop the time taken to $${\displaystyle \frac{1}{1000}}$$ of the original amplitude.
Again consider the formula.
$$A=A_0{e}^{-\gamma t}$$
Take ln on both the sides of the above function.
$$\ln {\displaystyle \frac{A_0}{A}}=\gamma t$$
Substitute the given values in the above equation.
$$\ln 1000=\gamma t$$
Now substitute the value of the decay constant obtained in the above equation.
$$\ln 1000={\displaystyle \frac{\ln 2}{2}}t$$
Continue the further calculation.
$$\begin{array}{rl}& t=2\left({\displaystyle \frac{\ln 1000}{\ln 2}}\right)\\ & \Rightarrow t=19.931s\end{array}$$
The value of the time taken can be rounded off to the nearest value as, t = 20 s.
As the time it will take to drop to$${\displaystyle \frac{1}{1000}}$$of the original amplitude is close to 20 s.

So, the correct answer is “Option B”.

Note:
The usage of the formula is the main step in this calculation. The unit of the parameters should be taken care of. As, in the options, the time is represented in terms of the seconds, even, the time taken can be asked in terms of the other units like minutes.