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Question
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
Options
E/2
2E
E
E/4
Solution
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 : \[E' = \frac{1}{2} k_B {x_B}^2 = \frac{1}{2}(\frac{k_A}{2})(2 x_A )^2 \]
\[ \therefore E' = 2 \times \left( \frac{1}{2} k_A {x_A}^2 \right) = 2E [\text{ From }(iii)]\]
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