In figure (A), mass $2m$ is fixed on mass $m$ which is attached to two springs of spring constant $k$. In figure (B), mass $m$ is attached to two springs of spring constant $k$ and $2k$. If mass $m$ in (A) and (B) are displaced by distance $x$ horizontally and then released, then the time periods $T_1$ and $T_2$ corresponding to (A) and (B) respectively follow the relation
Answer: (A) $\dfrac{T_1}{T_2} = \dfrac{3}{\sqrt2}$
Springs on both sides of the block act in parallel.
(A): mass $3m$, $k_{\text{eff}} = 2k$, so $T_1 = 2\pi\sqrt{\dfrac{3m}{2k}}$. (B): mass $m$, $k_{\text{eff}} = 3k$, so $T_2 = 2\pi\sqrt{\dfrac{m}{3k}}$.
$$\frac{T_1}{T_2} = \sqrt{\frac{3m}{2k}\cdot\frac{3k}{m}} = \sqrt{\frac92} = \frac{3}{\sqrt2}$$
Solution by Sreeraj P, M.Sc Physics