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