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. $2 \pi \sqrt{\frac{4 M _{1}+ M _{2}}{k}}$
• B. $2 \pi \sqrt{\frac{4 M_{2}+M_{1}}{k}}$
• C. $4 \pi \sqrt{\frac{2 M_{1}+M_{2}}{k}}$
• D. $\pi \sqrt{\frac{4 M _{1}- M _{2}}{ k }}$

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
$-M_{1} g x+\frac{1}{2} M_{1} v^{2}+M_{2} g \frac{x}{2}+\frac{1}{2} M_{2} \frac{v^{2}}{2}+\frac{1}{2} k \frac{x}{2}^{2}=$ const.
$-M_{1} g x+\frac{1}{2} M_{1} v^{2}+\frac{M_{2}}{2} g x+\frac{1}{8} M_{2} v^{2}+\frac{1}{8} k x^{2}=$ const.
Differentiate wrt time
$- M _{1} gv + M _{1} v \frac{ dv }{ dt }+\frac{ M _{2 g }}{2} v +\frac{1}{4} M v \frac{ dv }{ dt }+\frac{1}{8} kx.v =0 $
$\Rightarrow - M _{1} g + M _{1} \frac{ dv }{ dt }+\frac{ M _{2} g }{2}+\frac{1}{4} M _{2} \frac{ dv }{ dt }+\frac{1}{8} kx =0$

$M_{1}+\frac{M_{2}}{4} \frac{d v}{d t}=-\frac{k}{4} x+M_{1} g-\frac{M_{2} g}{2}$
$a=\frac{d v}{d t}=-\frac{k}{4 M_{1}+M_{2}} x+\frac{22 M_{1}-M_{2 g}}{4 M_{1}+M_{2}}$
Here, $ \omega^{2}=\frac{ k }{4 M _{1}+ M _{2}} \Rightarrow T =\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{4 M _{1}+ M _{2}}{ k }}$