Consider a single degree of freedom system comprising a mass M, supported on a spring and a dashpot as shown in the figure.

If the amplitude of the free vibration response reduces from 8 mm to 1.5 mm in 3 cycles, the damping ratio of the system is______ (round off to three decimal places).

The correct answer is 0.088.

Understanding the System:
The provided schematic diagram illustrates a standard single degree of freedom (SDOF) damped mass-spring system. The system consists of a mass block labeled M, which is supported on a spring labeled Spring and a damper labeled Dashpot connected in parallel to a fixed ground. When such a system is perturbed, it undergoes free underdamped vibrations, where the amplitude of vibration decays exponentially over time due to energy dissipation by the dashpot.

Mathematical Formulation:
The decay of amplitude in successive cycles of a free vibration response is characterized by the logarithmic decrement, denoted by $\delta$. The logarithmic decrement is defined as the natural logarithm of the ratio of any two successive amplitudes:

$\delta =\ln \left(\frac{x_k}{x_{k+1}}\right)$

For a decay occurring over $n$ full cycles, the ratio of the initial amplitude $x_0$ to the amplitude after $n$ cycles $x_n$ is given by the product of ratios of successive amplitudes:

$\frac{x_0}{x_n}=\frac{x_0}{x_1}\cdot \frac{x_1}{x_2}\cdot ...\cdot \frac{x_{n-1}}{x_n}=e^{n\delta }$

Taking the natural logarithm of both sides yields:

$n\delta =\ln \left(\frac{x_0}{x_n}\right)$

Step-by-Step Calculation:
From the problem statement, we have the following parameters:
- Number of cycles, $n=3$
- Initial amplitude, $x_0=8\text{mm}$
- Amplitude after 3 cycles, $x_3=1.5\text{mm}$

Substituting these values into the expression:

$\frac{x_0}{x_3}=\frac{8}{1.5}=\frac{80}{15}$

Thus:

$e^{\delta }\cdot e^{\delta }\cdot e^{\delta }=\frac{80}{15}\Rightarrow e^{3\delta }=\frac{80}{15}$

Taking the natural logarithm:

$3\delta =\ln \left(\frac{80}{15}\right)\Rightarrow 3\delta =1.67398\Rightarrow \delta =0.55799$

The logarithmic decrement $\delta$ is related to the damping ratio $\xi$ by the formula:

$\delta =\frac{2\pi \xi }{\sqrt{1-{\xi }^2}}$

Substituting $\delta =0.55799$ into the equation:

$\frac{2\pi \xi }{\sqrt{1-{\xi }^2}}=0.55799$

Squaring both sides to solve for $\xi$:

$\frac{4{\pi }^2{\xi }^2}{1-{\xi }^2}={\left(0.55799\right)}^2\approx 0.31135$

Rearranging the terms:

$4{\pi }^2{\xi }^2=0.31135\left(1-{\xi }^2\right)$

$\left(4{\pi }^2+0.31135\right){\xi }^2=0.31135$

Using $4{\pi }^2\approx 39.4784$:

$39.78977{\xi }^2=0.31135$

${\xi }^2=\frac{0.31135}{39.78977}\approx 0.007825$

$\xi =\sqrt{0.007825}\approx 0.08846$

Rounding to three decimal places, the damping ratio of the system is 0.088.