Question

Find the elastic potential energy stored in each spring shown in figure (12−E8), when the block is in equilibrium. Also find the time period of vertical oscillation of the block.
Figure

Answer 1

Explanation:

• We know that force = weight = mg.  So, at spring $\mathrm{k}_{1 \mathrm{c}} \mathrm{x}_{1}=\frac{\mathrm{m} \mathrm{g}}{\mathrm{k}_{1}}$, where m is mass and g is acceleration due to gravity.  
• When the block is in equilibrium, the elastic potential energy stored in spring $k_1$ is acceleration due to gravity.  When the block is in equilibrium, the elastic potential energy stored in spring $k_1$ is $\mathrm{PE}_{1}=(1 / 2) \mathrm{k}_{1} \mathrm{x}_{1}^{2}=\frac{\mathrm{m}^{2} \mathrm{g}^{2}}{2 \mathrm{k}_{1}}$. Likewise, $\mathrm{PE}_{2}=\frac{\mathrm{m}^{2} \mathrm{e}^{2}}{2 \mathrm{k}_{2}}$ and $\mathrm{PE}_{3}=\frac{\mathrm{m}^{2} \mathrm{s}^{2}}{2 \mathrm{k}_{3}}$.

• We know that the time period of vertical oscillation of the block  $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$, where $\mathrm{k}=\frac{\mathrm{k}_{1} \mathrm{k}_{2} \mathrm{k}_{\mathrm{a}}}{\mathrm{k}_{1} \mathrm{k}_{2}+\mathrm{k}_{2} \mathrm{k}_{\mathrm{a}}+\mathrm{k}_{\mathrm{a}} \mathrm{k}_{\mathrm{l}}}$. Thus, on substituting the value, we get the time period $\mathrm{T}=2 \pi \sqrt{m\left(\frac{1}{k_{1}}+\frac{1}{k_{2}}+\frac{1}{k_{3}}\right)}.$