A block of mass m is attached to two springs | vedclass

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(B) The effective spring constant $k_{eff}$ for two springs in parallel is $k_{eff} = k_1 + k_2$.Using the principle of conservation of energy,the tot

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$\sqrt{\frac{3(k_1 + k_2)x^2}{4m}}$

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(B) The effective spring constant $k_{eff}$ for two springs in parallel is $k_{eff} = k_1 + k_2$.Using the principle of conservation of energy,the total mechanical energy at the extreme position $(x)$ is equal to the total mechanical energy at position $x/2$.At the extreme position $(x)$,the velocity is zero,so the total energy is purely potential: $E = \frac{1}{2} k_{eff} x^2 = \frac{1}{2} (k_1 + k_2) x^2$.At position $x/2$,the total energy is the sum of potential and kinetic energy: $E = \frac{1}{2} k_{eff} (x/2)^2 + \frac{1}{2} m v^2$.Equating the energies: $\frac{1}{2} (k_1 + k_2) x^2 = \frac{1}{2} (k_1 + k_2) (x/2)^2 + \frac{1}{2} m v^2$.$(k_1 + k_2) x^2 = (k_1 + k_2) \frac{x^2}{4} + m v^2$.$m v^2 = (k_1 + k_2) x^2 - \frac{(k_1 + k_2) x^2}{4} = \frac{3}{4} (k_1 + k_2) x^2$.$v^2 = \frac{3(k_1 + k_2) x^2}{4m}$.$v = \sqrt{\frac{3(k_1 + k_2) x^2}{4m}}$.

$A$ block of mass $m$ is attached to two springs of spring constants $k_1$ and $k_2$ as shown in the figure. The block is displaced by $x$ towards the right and released. The velocity of the block when it is at $x/2$ will be

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