A mass m attached to free end of a spring executes SHM with a...

Time period of oscillation is given by $T = 2\pi\sqrt{\dfrac{m}{k}}$, where $k$ is the force constant of the spring.

$∵$ Spring is same, therefore, spring constant will be same.

Putting the different values,

In the first case $T = 1 = 2\pi\sqrt{\dfrac{m}{k}}$

In the second case ${T}^{'} = 2 = 2\pi\sqrt{\dfrac{m + 3}{k}}$

Dividing the equations, we get

$\dfrac{T}{{T}^{'}} = \dfrac{1}{2} = \sqrt{\dfrac{m}{m + 3}}$

$⇒ \dfrac{m}{m + 3} = \dfrac{1}{4} ⇒ 4m = m + 3 ⇒ m = 1$

Thus, $m = 1kg$