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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2013
  • Hard
  • MCQ

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

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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 \(\mathrm{B}\)

\(x_{B}=(8.75-x) c m\)

\(\mathrm{F}=300 \times x=400(8.75-x)\)

Solving we get, \(x=5 \mathrm{cm}\)

\(x_{B}=8.75-5=3.75 \mathrm{cm}\)

\(\frac{E_{A}}{E_{B}}=\frac{\frac{1}{2} k_{A}\left(x_{A}\right)^{2}}{\frac{1}{2} k_{B}\left(x_{B}\right)^{2}}=\frac{300 \times(5)^{2}}{400 \times(3.75)^{2}}=\frac{4}{3}\)

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