Two springs A and B(kA = 2kB) are stretched by applying forces of equal magnitudes a the four ends. If the energy stored in A is E, that in B isA

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2E Let xA and xB be the extensions produced in springs A and B, respectively. Restoring force on spring A, F = k_A x_A ...(i) Restoring force on spring B, F = k_B x_B ...(ii) From (i) and (ii), we get: k_A x_A = k_B x_B It is given that kA = 2kB \therefore x_B = 2 x_A Energy stored in spring A : E = $\frac{1}{2}$ k_A {x_A}^2 ...(iii) Energy stored in spring B : E' = $\frac{1}{2}$ k_B {x_B}^2 = $\frac{1}{2}$(k_A/2)(2 x_A )^2 \therefore E' = 2 $\times$ \left( $\frac{1}{2}$ k_A {x_A}^2 \right) = 2E [\text{ From }(iii)]

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