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A trolley of mass m is connected to two identical springs, each of force constant k, as shown in fig. The trolly is displaced from its equilibrium position by a distance x and released. The trolly executes simple harmonic motion of period T. After some time it comes to rest due to friction. The total energy dissipated as heat is (assume the damping force to be weak)

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$\frac{4 π^{2} {mx}^{2}}{T^{2}}$

$\frac{{mx}^{2}}{T^{2}}$

$\frac{1}{2} {kx}^{2}$

${kx}^{2}$

answer is B.

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Detailed Solution

Two springs are connected in parallel equivalent spring constant is – 2K.
Total energy $\frac{1}{2} (2 K) x^{2} = {kx}^{2}$
The final energy of the system = 0
Hence the total energy dissipated is = ${kx}^{2}$

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