Question:
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:
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: $k_{\text{eq}} = k_1 + k_2$.
Add the spring constants directly: $k_{\text{eq}} = k_1 + k_2$.
Updated On: Jul 22, 2026
- $2\pi \sqrt{\frac{m}{k}}$
- $2\pi \sqrt{\frac{m}{2k}}$
- $2\pi \sqrt{\frac{2m}{k}}$
- $2\pi \sqrt{\frac{3m}{2k}}$
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
We need to find the time period of oscillation for a mass connected between two identical springs attached to fixed walls.
Step 2: Key Formula and Approach:
The time period $T$ of a mass-spring system is:
\[ T = 2\pi \sqrt{\frac{m}{k_{\text{eq}}}} \] where $k_{\text{eq}}$ is the equivalent spring constant.
When a mass is suspended between two springs, any displacement of the mass stretches one spring and compresses the other.
Both springs exert restoring forces in the same direction, which means they are in a parallel configuration.
Step 3: Detailed Explanation:
• Calculate equivalent spring constant ($k_{\text{eq}}$):
For springs connected in parallel:
\[ k_{\text{eq}} = k_1 + k_2 \] Given $k_1 = k_2 = \frac{k}{3}$:
\[ k_{\text{eq}} = \frac{k}{3} + \frac{k}{3} = \frac{2k}{3} \]
• Calculate the time period ($T$):
\[ T = 2\pi \sqrt{\frac{m}{k_{\text{eq}}}} \] Substitute $k_{\text{eq}}$:
\[ T = 2\pi \sqrt{\frac{m}{\frac{2k}{3}}} = 2\pi \sqrt{\frac{3m}{2k}} \]
Step 4: Final Answer:
The time period of oscillation is $2\pi \sqrt{\frac{3m}{2k}}$, which corresponds to Option (D).
We need to find the time period of oscillation for a mass connected between two identical springs attached to fixed walls.
Step 2: Key Formula and Approach:
The time period $T$ of a mass-spring system is:
\[ T = 2\pi \sqrt{\frac{m}{k_{\text{eq}}}} \] where $k_{\text{eq}}$ is the equivalent spring constant.
When a mass is suspended between two springs, any displacement of the mass stretches one spring and compresses the other.
Both springs exert restoring forces in the same direction, which means they are in a parallel configuration.
Step 3: Detailed Explanation:
• Calculate equivalent spring constant ($k_{\text{eq}}$):
For springs connected in parallel:
\[ k_{\text{eq}} = k_1 + k_2 \] Given $k_1 = k_2 = \frac{k}{3}$:
\[ k_{\text{eq}} = \frac{k}{3} + \frac{k}{3} = \frac{2k}{3} \]
• Calculate the time period ($T$):
\[ T = 2\pi \sqrt{\frac{m}{k_{\text{eq}}}} \] Substitute $k_{\text{eq}}$:
\[ T = 2\pi \sqrt{\frac{m}{\frac{2k}{3}}} = 2\pi \sqrt{\frac{3m}{2k}} \]
Step 4: Final Answer:
The time period of oscillation is $2\pi \sqrt{\frac{3m}{2k}}$, which corresponds to Option (D).
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