Question

Figure shows a system consisting of a massless pulley, a spring of force constant $k$ and a block of mass $m$. If the block is slightly displaced vertically down from its equilibrium position and then released, the period of its vertical oscillation is

• A. $2\pi \sqrt{\frac{m}{k}}$
• B. $\pi \sqrt{\frac{m}{4k}}$
• C. $\pi \sqrt{\frac{m}{k}}$
• D. $4\pi \sqrt{\frac{m}{k}}$

Answer 1

The correct option is D $4\pi \sqrt{\frac{m}{k}}$

Let us assume that in equilibrium condition, spring is $x_0$ elongated from its natural length.


In equilibrium, $T_{\circ }=mg$
and $k{x}_{\circ }=2T_{\circ }\Rightarrow k{x}_{\circ }=2mg\left(i\right)$
If the mass $m$ moves down a distance $x$ from its equilibrium postion, then pulley will move down by $\frac{x}{2}$. So the extra force in spring will be $\frac{kx}{2}$. From figure, net force on block
$F_{net}=mg-T-\frac{k}{2}\left(x_{\circ }+\frac{x}{2}\right)$
Where, $2T=k(x_0+\frac{x}{2})$ (From net force on pulley)
$\Rightarrow F_{net}=mg-\frac{k{x}_{\circ }}{2}-\frac{kx}{4}$

From eq. (i)
$F_{net}=\frac{-kx}{4}$
Now compare eq. (ii) with $F=-K_{SHM}x,then$
$K_{SHM}=\frac{k}{4}\Rightarrow T=2\pi \sqrt{\frac{m}{K_{S.H.M}}}=2\pi \sqrt{\frac{4m}{k}}$