Question

A block of mass $M_1$ is constrained to move along with a moveable pulley of mass $M_2$ which is connected to a spring of force constant $k$ , as shown in the figure. If the mass of the fixed pulley is negligible and friction is absent everywhere, then the period of small oscillations of the system is

• A. $2\pi \sqrt{\frac{4M_1+M_2}{k}}$
• B. $2\pi \sqrt{\frac{4M_2+M_1}{k}}$
• C. $4\pi \sqrt{\frac{2M_1+M_2}{k}}$
• D. $\pi \sqrt{\frac{4M_1-M_2}{k}}$

Answer 1

Answer: (A)

If the mass $M_1$ is displaced by $x$ and has speed $v$, spring will extend by $x/2$ and $M_2$ will have speed $v/2$. Energy of this system can be written as
$-M_{1}gx+\frac{1}{2}{M}_1{v}^2+M_{2}g\frac{x}{2}+\frac{1}{2}{M}_2\frac{v^2}{2}+\frac{1}{2}k{\frac{x}{2}}^2=$ const.
$-M_{1}gx+\frac{1}{2}{M}_1{v}^2+\frac{M_2}{2}gx+\frac{1}{8}{M}_2{v}^2+\frac{1}{8}k{x}^2=$ const.
Differentiate wrt time
$-M_{1}gv+M_{1}v\frac{dv}{dt}+\frac{M_{2g}}{2}v+\frac{1}{4}Mv\frac{dv}{dt}+\frac{1}{8}kx.v=0$
$\Rightarrow -M_{1}g+M_1\frac{dv}{dt}+\frac{M_{2}g}{2}+\frac{1}{4}{M}_2\frac{dv}{dt}+\frac{1}{8}kx=0$

$M_1+\frac{M_2}{4}\frac{dv}{dt}=-\frac{k}{4}x+M_{1}g-\frac{M_{2}g}{2}$
$a=\frac{dv}{dt}=-\frac{k}{4M_1+M_2}x+\frac{22M_1-M_{2g}}{4M_1+M_2}$
Here, ${\omega }^2=\frac{k}{4M_1+M_2}\Rightarrow T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{4M_1+M_2}{k}}$