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Logarithmic decrement of a damped single degree of freedom system is δ. If stiffness of the spring is doubled and mass is made half, then logarithmic decrement of the new system will be equal to


1. 1/2δ
2. δ
3. 2δ
4. 1/4 δ

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Correct Answer - Option 2 : δ

Concept:

Logarithmic decrement of spring is,

\(\delta = \frac{{2\pi\xi }}{{\sqrt {1 -\xi 2\;} }}\)

where, \(\xi = \frac{C}{{{C_c}}}\)

And CC = 2 mωn

The natural frequency of spring is, \({\omega _n} = \sqrt {\frac{k}{m}} \) 

When the stiffness of the spring is doubled and mass is made half,

The natural frequency of spring become \({\omega _n} = \sqrt {\frac{2k}{{\frac{m}{2}}}} = 2\sqrt {\frac{k}{m}} \) 

Now, \({\xi_1} = \frac{C}{{{C_C}}} = \frac{C}{{2m{\omega _n}}} =\xi\)

\({\xi_2} = \frac{C}{{{C_C}}} = \frac{C}{{2(m/2){2\omega _n}}} = \xi\)

\({\delta _2} = \frac{{2\pi\epsilon {_2}}}{{\sqrt {1 -\epsilon _2^2} }}\)

\( = \frac{{2\pi\epsilon }}{{\sqrt {1 - {\epsilon^2}} }}\)        ---- (1)

\(\delta = \frac{{2\pi\epsilon }}{{\sqrt {1 - {\epsilon^2}} }}\)       ---- (2)

Form (1) and (2)

\(\frac{{{\delta _2}}}{\delta } = \frac{{2\pi\epsilon }}{{2\pi\epsilon }}\frac{{\sqrt {1 - {\epsilon^2}} \;}}{{\sqrt {1 - {\epsilon^2}} }}\)

\(\frac{{{\delta _2}}}{\delta } = 1\)

⇒ δ2 = δ

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