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

Answer 1

Let us assume that in equilibrium condition, spring is $x_0$ elongated from its natural length.



In equilibrium, $T_0=mg$

and $k{x}_0=2T_0$

$\Rightarrow k{x}_0=2mg....\left(i\right)$

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,

$F_{net}=mg-T=mg-\frac{k}{2}(x_0+\frac{x}{2})$

$F_{net}=mg-\frac{k{x}_0}{2}-\frac{kx}{4}$

From eq. (i)

$F_{net}=\frac{-kx}{4}....\left(ii\right)$

Now compare eq. (ii) with

$F=-K_{SHM}x$ then

$K_{SHM}=\frac{k}{4}$

$T=2\pi \sqrt{\frac{m}{K_{SHM}}}=2\pi \sqrt{\frac{4m}{k}}$

$T=4\pi \sqrt{\frac{m}{k}}$

Final answer: (d)