Question

A spring has a certain mass suspended from it and its time period for vertical oscillation is T. The spring is now cut into two equal halves and same mass is suspended from one of the halves. The period of vertical oscillation is now

• A. $\raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
• B. $\raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.$
• C. $2T$
• D. $\sqrt{2}T$

Answer 1

The correct option is B

$\raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.$

1. Let us consider spring of length l and spring constant k as a combination of two identical springs of length $\raisebox{1ex}{$l$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$and spring constant k'.
2. From Hooke's law, $$\begin{gathered} k=\frac{k\text{'}k\text{'}}{k\text{'}+k\text{'}}=\frac{k\text{'}k\text{'}}{2k\text{'}}=\frac{k\text{'}}{2} \\ \Rightarrow k\text{'}=2k. \end{gathered}$$
3. In the first case, the time period of vertical oscillation, $T=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{k}}$(m is mass suspended from spring).
4. In the second case, the time period of vertical oscillation $$\begin{gathered} T\text{'}=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}\text{'}}}=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{2\mathrm{k}}}=\frac{1}{\sqrt{2}}\times 2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}}} \\ \Rightarrow \mathrm{T}\text{'}=\frac{\mathrm{T}}{\sqrt{2}} \end{gathered}$$

Hence, the correct option is B.