Two identical springs of each having force constant $\frac{k}{3}$ are connected as shown in the figure. If it executes simple harmonic motion, its time period is:
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A mass placed between two springs connected to opposite walls is always a parallel system.
Add the spring constants directly: $k_{\text{eq}} = k_1 + k_2$.
Step 1: Think in terms of stored energy instead of forces. When the mass is displaced by $x$, one spring stretches by $x$ and the other compresses by $x$, so both store elastic potential energy, $\frac{1}{2}k_1x^2$ and $\frac{1}{2}k_2x^2$ respectively. Step 2: Add the energies. The total potential energy of the system is \[ U = \frac{1}{2}k_1x^2 + \frac{1}{2}k_2x^2 = \frac{1}{2}(k_1+k_2)x^2 \] which has exactly the form for a single spring of effective stiffness $k_{\text{eq}} = k_1+k_2$, confirming the two springs act in parallel. Step 3: Substitute the given values. With $k_1=k_2=\frac{k}{3}$, $k_{\text{eq}} = \frac{2k}{3}$. Step 4: Get the time period. \[ T = 2\pi\sqrt{\frac{m}{k_{\text{eq}}}} = 2\pi\sqrt{\frac{3m}{2k}} \] \[ \boxed{T = 2\pi\sqrt{\frac{3m}{2k}}} \]
