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

Answer: (D)

Let us assume that in equilibrium condition spring is $x_0$ elongate from its natural length

In equilibrium $T_0=mg$
and $k{x}_0=2T_0$
$\Rightarrow k{x}_0=2mg$ (1)
If the mass $m$ moves down a distance $x$ from its equilibrium position then pulley will move down by
$\frac{x}{2}.$ So the extra force in spring will be $\frac{kx}{2}$. From figure
$F_{\text{net }}=mg-T=mg-\frac{k}{2}\left(x_0+\frac{x}{2}\right)$
$F_{\text{net }}=mg-\frac{k{x}_0}{2}-\frac{kx}{4}$
from eq. (1), we get
$F_{\text{net }}=\frac{-kx}{4}$ (2)
Now compare eq. (2) with $F=-K_{\text{sHM }}x$
then $K_{\text{SHM }}=\frac{K}{4}$
$\Rightarrow T=2\pi \sqrt{\frac{m}{K_{\text{SHM }}}}=2\pi \sqrt{\frac{4m}{K}}=4\pi \sqrt{\frac{m}{K}}$