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. $\pi \sqrt{\frac{m}{4 K}}$
• 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 $T_{0}=m g$
and $k x_{0}=2 T_{0}$
$\Rightarrow k x_{0}=2 m g$ (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 $\frac{k x}{2}$. From figure
$F_{\text {net }}=m g-T=m g-\frac{k}{2}\left(x_{0}+\frac{x}{2}\right)$
$F_{\text {net }}=m g-\frac{k x_{0}}{2}-\frac{k x}{4}$
from eq. (1), we get
$F_{\text {net }}=\frac{-k x}{4}$ (2)
Now compare eq. (2) with $F=-K_{\text {sHM }} x$
then $K_{\text {SHM }}=\frac{K}{4}$
$\Rightarrow T=2 \pi \sqrt{\frac{m}{K_{\text {SHM }}}}=2 \pi \sqrt{\frac{4 m}{K}}=4 \pi \sqrt{\frac{m}{K}}$