Time Period of Cut Springs in Parallel – Rankers Physics

Spring Block System: Practice Problem & Solution

When a block is suspended from a spring, time period of its oscillation is \(T\). If this spring is cut into 3 equal parts and this block is suspended from parallel combination of these 3 parts, then new time period of oscillation will be:
\(\sqrt{3}T\)
\(\frac{T}{\sqrt{3}}\)
3T
\(\frac{T}{3}\)

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Spring Block System. Understanding the underlying formula is key to arriving at the correct answer below:

Cutting a spring of constant \(k\) into 3 equal parts increases each part's spring constant to \(3k\). In a parallel connection, the equivalent constant is \(k_{\text{eq}} = 3k + 3k + 3k = 9k\). The new time period is \(T' = 2\pi\sqrt{\frac{m}{9k}} = \frac{T}{3}\).

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