Physics - SHM Question with Solution | TestHub

According to energy conservation principle, If, ${\mathrm{x}}_1$ is maximum elongation in the spring when the particle is in its lowest extreme position. Then, $$\begin{array}{l}\mathrm{mgh}=\frac{1}{2}{\mathrm{kx}}_1^2-{\mathrm{mgx}}_1\\ \Rightarrow \frac{1}{2}{\mathrm{kx}}_1^2-{\mathrm{mgx}}_1-\mathrm{mgh}=0\\ \text{ or, }{\mathrm{x}}_1^2-\frac{2\mathrm{mg}}{\mathrm{k}}{\mathrm{x}}_1-\frac{2\mathrm{mg}}{\mathrm{k}}\cdot \mathrm{h}=0\\ \therefore {\mathrm{x}}_1=\frac{\frac{2\mathrm{mg}}{\mathrm{k}}\pm \sqrt{{\left(\frac{2\mathrm{mg}}{\mathrm{k}}\right)}^2+4\times \frac{2\mathrm{mg}}{\mathrm{k}}\mathrm{h}]}}{2}\end{array}$$ Amplitude $\mathrm{A}={\mathrm{X}}_1-{\mathrm{X}}_0$ (elongation in spring for equilibrium position) $$\mathrm{A}=\frac{\mathrm{mg}}{\mathrm{k}}\sqrt{\left(1+\frac{2\mathrm{hk}}{\mathrm{mg}}\right)}$$