Question:

If two similar springs each of spring constant \(K_1\) are joined in series, the new spring constant and time period would be changed by a factor :

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For two identical springs in series, \(K_{eq} = K_1/2\); then use \(T = 2\pi\sqrt{m/K}\) to see how the period scales.
Updated On: Nov 4, 2026
  • \(\frac{1}{2}, 2\sqrt{2}\)
  • \(\frac{1}{4}, 2\sqrt{2}\)
  • \(\frac{1}{2}, \sqrt{2}\)
  • \(\frac{1}{4}, \sqrt{2}\)
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The Correct Option is C

Solution and Explanation

Springs in series: \(\frac{1}{K_{eq}} = \frac{1}{K_1}+\frac{1}{K_1} = \frac{2}{K_1}\), so \(K_{eq} = \frac{K_1}{2}\). The spring constant changes by a factor \(\frac{1}{2}\).
Time period \(T = 2\pi\sqrt{\frac{m}{K}}\) gives \(T \propto K^{-1/2}\).
\(\frac{T'}{T} = \sqrt{\frac{K_1}{K_1/2}} = \sqrt{2}\).
Hence the factors are \(\frac{1}{2}\) and \(\sqrt{2}\).
Answer: (3) \(\frac{1}{2}, \sqrt{2}\)
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