Question

A system shown in the figure consists of a massless pulley, a spring of force constant $k$ and a block of mass $m$ . If the block is slightly displaced vertically downwards from its equilibrium position and released, then the period of vertical oscillations is

• A. $\textit{T}=\text{\pi }\sqrt{\left(\frac{m }{\text{4} k}\right)}$
• B. $T=2\text{\pi }\sqrt{\left(\frac{m }{\text{4} k}\right)}$
• C. $T=2\text{\pi }\sqrt{\left(\frac{m }{\text{2} k}\right)}$
• D. $T=2\text{\pi }\sqrt{\left(\frac{m }{\text{3} k}\right)}$

Answer 1

Answer: (B)

In this situation, if mass m moves downwards a distance $x$ from the equilibrium position, the pulley will also move by x and so, the spring will stretch by $2x$ .
Therefore, the spring force will be $2kx$ . The restoring force on the block will be $4kx$ .
Hence, $F=-4kx$

$\Rightarrow \, m g - 4 k x = m a$
$\Rightarrow \, m a = - 4 k x + m g$
$\Rightarrow $ $a = - \frac{4 k x}{m }$ $+ g \Rightarrow \omega ^{2} = \left|\right. \frac{a}{x} \left|\right.$
$T=2\text{\pi }\sqrt{\left|\right. \frac{x}{a} \left|\right.}$
$T=2\text{\pi }\sqrt{\left(\right. \frac{m }{4 k} \left.\right)}$