d
\(\because A=A_{0} e^{-\frac{b t}{2 m}}\left(\text { where }, A_{0}=\text { maximum amplitude }\right)\)

\( According \,to \,the\, questions,\, after\, 5\, second, \)

\(0.9 \mathrm{A}_{0}=\mathrm{A}_{0} \mathrm{e}^{-\frac{\mathrm{b}(5)}{2 \mathrm{m}}}\)                        \(...(i)\)

\(After\, 10 \,more\, second,\)

\(A=A_{0} \mathrm{e}^{-\frac{\mathrm{b}(15)}{2 \mathrm{m}}}\)            \(...(ii)\)

From eq"s \((i)\) and \((ii)\) 

\(A=0.729 A_{0} \therefore \alpha=0.729\)