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

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=-4 kx

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