Question

Find the period of the free oscillations of the arrangement shown below, if mass $M_1$ is pulled down a little.
[Force constant of the spring is $k$ and mass of the fixed pulley is negligible]

• A. $T=2\pi \sqrt{\frac{M_1+M_2}{k}}$
• B. $T=2\pi \sqrt{\frac{M_1+4M_2}{k}}$
• C. $T=2\pi \sqrt{\frac{M_2+4M_1}{k}}$
• D. $T=2\pi \sqrt{\frac{M_2+3M_1}{k}}$

Answer 1

The correct option is C $T=2\pi \sqrt{\frac{M_2+4M_1}{k}}$

When the mass $m$ is pulled down through a small distance $x$ & released.
The energy of the system is given by
$\frac{1}{2}{M}_2{v}_2^2+\frac{1}{2}{M}_1{v}_1^2+\frac{1}{2}k{x}^2=\text{constant}.......\left(1\right)$
Where the displacement of the body of mass $M_2=x$ & that of mass $M_1=2x$ & the deformation of the spring is $x$.
From this we can deduce that ,speed of the mass $M_2$
$\frac{dx}{dt}=v_2=\frac{v_1}{2}......\left(2\right)$
Using $\left(2\right)$ in $\left(1\right)$ we get,
$\frac{1}{2}{M}_2{v}_2^2+\frac{1}{2}{m}_1(2v_2{)}^2+\frac{1}{2}k{x}^2=\text{constant}$
Differentiating both sides with respect to $t$, we get
$\frac{1}{2}(M_2+4M_1)\left(2v_2\frac{d{v}_2}{dt}\right)+\frac{1}{2}k2x.\frac{dx}{dt}=0$
$\Rightarrow (M_2+4M_1)\frac{d{v}_2}{dt}=-kx[\because v_2\ne 0]$
$\Rightarrow \frac{d^{2}x}{d{t}^2}=-\left(\frac{k}{4M_1+M_2}\right)x$
Comparing this with $a=-{\omega }^{2}x$ we get,
$\omega =\sqrt{\frac{k}{(4M_1+M_2)}}$
Time period of oscillation $T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{4M_1+M_2}{k}}$
Hence, option (c) is the correct answer.