Question:
The time period of a mass suspended from a spring is \( T \). If the spring is cut into three equal parts and connected in parallel. The same mass is suspended from these parallel springs, then the new time period of the mass will be
The time period of a mass suspended from a spring is \( T \). If the spring is cut into three equal parts and connected in parallel. The same mass is suspended from these parallel springs, then the new time period of the mass will be
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Cutting spring $\longrightarrow$ stiffness increases. Parallel $\longrightarrow$ stiffness adds. Always track \(k\) carefully!
Updated On: Apr 22, 2026
- \( \frac{T}{4} \)
- \( T \)
- \( \frac{T}{3} \)
- \( 3T \)
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The Correct Option is C
Solution and Explanation
Concept:
Time period of a spring-mass system:
\[
T = 2\pi \sqrt{\frac{m}{k}}
\]
If a spring is cut into \( n \) equal parts, each part has spring constant \( nk \).
For springs in parallel:
\[
k_{\text{eq}} = k_1 + k_2 + \cdots
\]
Step 1: Spring cut into 3 parts.
Each part has spring constant: \[ k' = 3k \]
Step 2: Parallel combination.
\[ k_{\text{eq}} = 3k + 3k + 3k = 9k \]
Step 3: New time period.
\[ T' = 2\pi \sqrt{\frac{m}{9k}} = \frac{1}{3} \cdot 2\pi \sqrt{\frac{m}{k}} = \frac{T}{3} \]
Step 1: Spring cut into 3 parts.
Each part has spring constant: \[ k' = 3k \]
Step 2: Parallel combination.
\[ k_{\text{eq}} = 3k + 3k + 3k = 9k \]
Step 3: New time period.
\[ T' = 2\pi \sqrt{\frac{m}{9k}} = \frac{1}{3} \cdot 2\pi \sqrt{\frac{m}{k}} = \frac{T}{3} \]
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