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

  $\mathit{\text{T}}=\text{π}\sqrt{\left(\frac{m}{\text{4}k}\right)}$
• Correct answer
  B

  $T=2\text{π}\sqrt{\left(\frac{m}{\text{4}k}\right)}$
• C

  $T=2\text{π}\sqrt{\left(\frac{m}{\text{2}k}\right)}$
• D

  $T=2\text{π}\sqrt{\left(\frac{m}{\text{3}k}\right)}$

Solution

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 $4 kx$.

Hence, $F=- 4 kx$



$\Rightarrow mg-4kx=ma$

$\Rightarrow ma=-4kx+mg$

$\Rightarrow$    $a=-\frac{4kx}{m}$ $+g\Rightarrow {\omega }^2=\left|\frac{a}{x}\right|$

                $T=2\text{π}\sqrt{\left|\frac{x}{a}\right|}$

                $T=2\text{π}\sqrt{\left(\frac{m}{4k}\right)}$

Hence, the correct answer is option: 'B'.