Question

In the arrangement shown in the diagram, pulleys are small and springs are ideal. $k_{1}=k_{2}=k_{3}=k_{4}=10 \, N \, m^{- 1}$ are force constants of the springs and mass $m=10 \, kg$ . If the time period of small vertical oscillations of the block of mass $m$ is given by $2\pi \text{x seconds}$ , then find the value of $x$ .

Answer 1

$\textit{T}=2\pi \sqrt{\frac{\textit{m}}{\textit{K}_{\text{eq}}}}$
where $\textit{K}_{\text{eq}}=\frac{1}{4 \, \left[\frac{1}{\textit{k}_{1}} + \frac{1}{\textit{k}_{2}} + \frac{1}{\textit{k}_{3}} + \frac{1}{\textit{k}_{4}}\right]}=\frac{1 0}{1 6}$
$\textit{T}=2\pi \sqrt{\frac{\textit{m}}{\textit{K}_{\text{eq}}}}=2\pi \text{x}$
So, $\text{x} = \sqrt{\frac{1 0}{1 0 / 1 6}} = 4 \text{, } \text{x} = 4$