A spring has a certain mass suspended from it and its period of vertical oscillations is $\mathrm{T}_1$. The…
A spring has a certain mass suspended from it and its period of vertical oscillations is $\mathrm{T}_1$. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now $\mathrm{T}_2$. The ratio of $T_2 / T_1$ is
$1: 2$
$1: \sqrt{2}$
$\sqrt{2}: 1$
$2: 1$
Solution
$\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{~m}}{\mathrm{~K}}} \Rightarrow \mathrm{~T} \propto \frac{1}{\sqrt{\mathrm{~K}}} \\
\therefore \quad & \frac{\mathrm{~T}_2}{\mathrm{~T}_1}=\sqrt{\frac{\mathrm{K}_1}{\mathrm{~K}_2}}
\end{aligned}...(i)$
Let K be the spring constant of the spring cut in half.
For series combination,
$\mathrm{K}_1=\frac{\mathrm{K} \times \mathrm{K}}{\mathrm{~K}+\mathrm{K}}=\frac{\mathrm{K}}{2}$ For second case, the mass is hung to only one half of the spring,
$\mathrm{K}_2=\mathrm{K}$ Substituting in $\mathrm{K}_1$ and $\mathrm{K}_2$ in (i),
$\frac{T_2}{T_1}=\sqrt{\frac{K / 2}{K}}=\frac{1}{\sqrt{2}}$