Question: In a damped oscillator with, m = 250 g, k = N/m and b = 70 g / s, what is the ratio of the amplitude of the damped oscillations to the initial amplitude at the end of 20cycles?

1. The mass of the object is,$m=250\text{g}=0.25\text{kg}$
2. The damping constant of the oscillator is,$b=70\text{g}/\text{s}=0.070\text{kg}/\text{s}$
3. Spring constant,$k=85\text{N}/\text{m}$

Using the equation for amplitude of forced oscillation, we can find the amplitude of oscillation after 20 cycles. Then we can find its ratio with initial amplitude.

The amplitude of oscillations after n cycles is given as-

${\mathbf{x}}_{\mathbf{n}}\mathbf{=}{\mathbf{x}}_{\mathbf{m}}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{bt}}{\mathbf{2}\mathbf{m}}}$

The time period of oscillation in the case of spring is given as-

$\mathbf{T}\mathbf{=}\mathbf{2}\mathbf{\pi }\sqrt{\frac{\mathbf{m}}{\mathbf{k}}}$

Here, m is the mass of the pendulum, k is the force constant of the spring, b is a damping constant and t is the time taken.

Suppose the period of oscillation is T and the initial amplitude is x_m

Now, we have equation for amplitude of the oscillation after n cycle is,

$x_n=x_m{e}^{-\frac{bt}{2m}}$

So, for n = 20, t = 20T, we get

$x_{20}=x_m{e}^{-\frac{b\times 20T}{2m}}$

The time period of oscillation can be expressed as

$$\begin{gathered} T=2\pi \sqrt{\frac{m}{k}} \\ =2\pi \sqrt{\frac{0.25\text{kg}}{85}} \\ =0.34\text{s} \end{gathered}$$

Using this value of period and the given values, the amplitude of oscillation after 20 cycles is

$$\begin{gathered} {\text{x}}_{20}={\text{x}}_m\times e^{-\left(0.070\text{kg}/\text{s}\right)\times 20\times \frac{0.34\text{s}}{2\times 0.25\text{kg}}} \\ \Rightarrow {\text{x}}_{20}={\text{x}}_m\times 0.39 \end{gathered}$$

Hence, the ratio of amplitudes is

$$\begin{gathered} \frac{{\text{x}}_{20}}{{\text{x}}_{\text{0}}}=\frac{{\text{x}}_m\times 0.39}{{\text{x}}_m} \\ =0.39 \end{gathered}$$

So, the required ratio of amplitudes is 0.39.