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JEE Mains · Physics · STD 11 - 13. oscillations

Two springs of force constants \(300\, N/m\) (Spring \(A\)) and \(400\, N/m\) (Spring \(B\)) are joined together in series . The combination is compressed by \(8.75\, cm\). The ratio of energy stored in \(A\) and \(B\) is  \(\frac{{{E_A}}}{{{E_B}}}\).  Then \(\frac{{{E_A}}}{{{E_B}}}\) is equal to

  1. A \(\frac{4}{3}\)
  2. B \(\frac{16}{9}\)
  3. C \(\frac{3}{4}\)
  4. D \(\frac{9}{16}\)
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Answer & Solution

Correct Answer

(A) \(\frac{4}{3}\)

Step-by-step Solution

Detailed explanation

\(\mathrm{k}_{\mathrm{A}}=300 \mathrm{N} / \mathrm{m}, \quad \mathrm{k}_{\mathrm{B}}=400 \mathrm{N} / \mathrm{m}\) Let when the combination of springs is compressed by force \(\mathrm{F}\). Spring \(A\) is compressed by \(x\). Therefore compression in spring \(\mathrm{B}\)…
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