time period \((T)=2 \pi\) sec.
Therefore time period \(T=2 \pi \times \sqrt{\frac{m}{k}} \Rightarrow \sqrt{\frac{5}{k}}=1\)
or \(k=5 N / m\).
According to Hooke's Law, \(F=-k l\).
Therefore decrease in length \((l)=-\frac{F}{k}=-\frac{5 g}{5}\) \(=-g\) metres