The springs are connected to the blocks as shown in figures A and B. When the blocks are slightly displaced and released, they oscillate with time periods T_A and T_B respectively. Then, the value of T_AT_B is:
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Whenever two springs act on a block from opposite sides, both forces act in the same restoring direction. So always treat them as parallel:
\[
K_{\text{eq}} = K_1 + K_2
\]
Step 1: Recall the time period formula. For a block attached to springs, the time period of oscillation is: \[ T = 2\pi \sqrt{\frac{m}{K_{eq}}} \] Here $m$ is the mass and $K_{eq}$ is the effective spring constant of the whole arrangement.
Step 2: Analyse system A. In system A the block of mass $3m$ has two identical springs of constant $K$, one on each side. When the block is pushed, both springs push it back together. Their effects add, so the effective constant is: \[ K_A = K + K = 2K \]
Step 3: Write the time period of A. Using the formula: \[ T_A = 2\pi \sqrt{\frac{3m}{2K}} \]
Step 4: Analyse system B. In system B the block of mass $m$ has springs that combine to give an effective constant of $3K$. Using the formula: \[ T_B = 2\pi \sqrt{\frac{m}{3K}} \]
Step 5: Take the ratio. Divide one period by the other so the $2\pi$ cancels: \[ \frac{T_A}{T_B} = \sqrt{\frac{3m/2K}{m/3K}} = \sqrt{\frac{3m}{2K}\cdot \frac{3K}{m}} \]
Step 6: Simplify the ratio. Inside the root, $m$ and $K$ cancel: \[ \frac{T_A}{T_B} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}} \] which matches the first option. \[ \boxed{\dfrac{3}{\sqrt{2}}} \]
