Physics - SHM Question with Solution | TestHub

If the mass${\mathrm{M}}_1$is displaced by$\mathrm{x}$and has speed$\mathrm{v}$, spring will extend by$\mathrm{x}/2$and${\mathrm{M}}_2$will have speed$\mathrm{v}/2$. Energy of this system can be written as

$-{\text{M}}_1\text{gx}+\frac{1}{2}{\text{M}}_1{\text{v}}^2+{\text{M}}_2\text{g}\frac{\text{x}}{2}+\frac{1}{2}{\text{M}}_2{\left(\frac{\text{v}}{2}\right)}^2+\frac{1}{2}\mathrm{k}{\left(\frac{\text{x}}{2}\right)}^2=\text{const.}$

$-{\text{M}}_1\text{gx}+\frac{1}{2}{\text{M}}_1{\text{v}}^2+\frac{{\text{M}}_2}{2}\text{gx}+\frac{1}{8}{\text{M}}_2{\text{v}}^2+\frac{1}{8}\mathrm{k}{\text{x}}^2=\text{const.}$

Differentiate wrt time

$-{\text{M}}_1\text{gv}+{\text{M}}_1\text{v}\frac{\text{dv}}{\text{dt}}+\frac{{\text{M}}_2\text{g}}{2}\text{v}+\frac{1}{4}{\text{M}}_2\text{v}\frac{\text{dv}}{\text{dt}}+\frac{1}{8}\mathrm{k}\text{x.v}=0$

$\Rightarrow -{\text{M}}_1\text{g}+{\text{M}}_1\frac{\text{dv}}{\text{dt}}+\frac{{\text{M}}_2\text{g}}{2}+\frac{1}{4}{\text{M}}_2\frac{\text{dv}}{\text{dt}}+\frac{1}{8}\mathrm{k}\text{x}=0$



$\left({\text{M}}_1+\frac{{\text{M}}_2}{4}\right)\frac{\text{dv}}{\text{dt}}=-\frac{\mathrm{k}}{4}\text{x}+{\text{M}}_1\text{g}-\frac{{\text{M}}_2\text{g}}{2}$

$\mathrm{a}=\frac{\text{dv}}{\text{dt}}=-\left(\frac{\mathrm{k}}{4{\text{M}}_1+{\text{M}}_2}\right)\text{x}+\frac{2\left(2{\text{M}}_1-{\text{M}}_2\right)\text{g}}{4{\text{M}}_1+{\text{M}}_2}$

Here,${\mathrm{\omega }}^2=\frac{\mathrm{k}}{4{\text{M}}_1+{\text{M}}_2}\Rightarrow \text{T}=\frac{2\mathrm{\pi }}{\mathrm{\omega }}=2\mathrm{\pi }\sqrt{\frac{4{\text{M}}_1+{\text{M}}_2}{\mathrm{k}}}$