All the springs in fig. (a), (b) and (c) are identical, each having force constant $K$ each. Mass $m$ is…
- $\mathrm{T}_{\mathrm{a}}=\sqrt{2} \mathrm{~T}_{\mathrm{b}}$
- $\mathrm{T}_{\mathrm{a}}=\frac{\mathrm{T}_{\mathrm{c}}}{\sqrt{2}}$
- $\mathrm{T}_{\mathrm{b}}=2 \mathrm{~T}_{\mathrm{a}}$
- $\mathrm{T}_{\mathrm{b}}=2 \mathrm{~T}_{\mathrm{c}}$
Solution
The time period for a spring-mass system is given by $T = 2\pi \sqrt{m / K_{\text{eq}}}$, where $m$ is the mass and $K_{\text{eq}}$ is the equivalent spring constant.
System (a): With one spring of constant $K$, $K_{\text{eq},a} = K$ and $T_a = 2\pi \sqrt{m / K}$.
System (b): Two springs in series yield $1 / K_{\text{eq},b} = 1 / K + 1 / K = 2 / K$, so $K_{\text{eq},b} = K / 2$ and $T_b = 2\pi \sqrt{2m / K} = \sqrt{2} \, T_a$.
System (c): Two springs in parallel yield $K_{\text{eq},c} = K + K = 2K$, so $T_c = 2\pi \sqrt{m / (2K)} = T_a / \sqrt{2}$.
Evaluating the options:
Option A: $T_a = \sqrt{2} \, T_b$ gives $T_a = \sqrt{2} (\sqrt{2} \, T_a) = 2 T_a$, which is false.
Option B: $T_a = T_c / \sqrt{2}$ gives $T_a = (T_a / \sqrt{2}) / \sqrt{2} = T_a / 2$, which is false.
Option C: $T_b = 2 T_a$ is false since $T_b = \sqrt{2} \, T_a$.
Option D: $T_b = 2 T_c$ gives $\sqrt{2} \, T_a = 2 (T_a / \sqrt{2}) = \sqrt{2} \, T_a$, which is true.
The correct relationship is $T_b = 2 T_c$.
Asked in: MHT CET 2025 (05 May Shift 2)