A mass m attached to a spring of spring constant k is stretched a distance x₀ from its equilibrium position and released with no initial velocity. The maximum speed attained by mass in its subsequent motion and the time at which this speed would be attained are, respectively,

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$\sqrt{\frac{k}{m}} x_{0}, π \sqrt{\frac{m}{k}}$

$\sqrt{\frac{k}{m}} \frac{x_{0}}{2}, \frac{π}{2} \sqrt{\frac{m}{k}}$

$\sqrt{\frac{k}{m}} x_{0}, \frac{π}{2} \sqrt{\frac{m}{k}}$

$\sqrt{\frac{k}{m}} \frac{x_{0}}{2}, π \sqrt{\frac{m}{k}}$

answer is C.

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Detailed Solution

At mean position, the speed will be maximum

$\frac{{kx}_{0}^{2}}{2} = \frac{{mv}^{2}}{2} \Rightarrow v_{max} = \sqrt{\frac{k}{m}} x_{0}$

and this is attained at $t = T / 4$

Time period of motion is $T = 2 π \sqrt{\frac{m}{k}}$

So required time is $t = \frac{π}{2} \sqrt{\frac{m}{k}}$

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