For a vibrating system, the successive amplitudes of vibration ob

SOLUTION

Concept:

Ratio of successive amplitude is given by

\(\frac{{{{\bf{X}}_0}}}{{{{\bf{X}}_1}}} = \frac{{{{\bf{X}}_1}}}{{{{\bf{X}}_2}}} = \frac{{{{\bf{X}}_2}}}{{{{\bf{X}}_3}}} = \ldots \ldots \ldots \ldots \ldots \ldots = \frac{{{{\bf{X}}_{\bf{n}}}}}{{{{\bf{X}}_{{\bf{n}} + 1}}}} = {{\bf{e}}^{\bf{δ }}}\)

where δ = Logarithmic amplitude

Logarithmic decrement is given by

\({\bf{δ }} = \frac{1}{{\bf{n}}}{\bf{ln}}\left( {\frac{{{{\bf{X}}_{\bf{o}}}}}{{{{\bf{X}}_{\bf{n}}}}}} \right)\)

Damping ratio(ζ) is given by

\(ζ = \frac{δ }{{\sqrt {4{\pi ^2}\; + \;{δ ^2}} }}\)

Calculation:

Given:

X₀ = 0.7, X₁ = 0.28, X₂ = 0.25, X₃ = 0.23, X₄ = 0.067, ζ = ?

Now, we know that

\({\bf{δ }} = \frac{1}{{\bf{n}}}{\bf{ln}}\left( {\frac{{{{\bf{X}}_{\bf{o}}}}}{{{{\bf{X}}_{\bf{n}}}}}} \right) = \frac{1}{4}ln\left( {\frac{{0.7}}{{0.067}}} \right)\)

∴ δ = 0.587

The damping ratio is given by

\(ζ = \frac{\delta }{{\sqrt {4{\pi ^2}\; + \;{\delta ^2}} }} = \;\frac{{0.587}}{{\sqrt {4{\pi ^2}\; + \;{{0.587}^2}} }}\)

∴ ζ = 0.0928