A particle of mass m moves in the potential energy U shown in the figure. Find the period of the motion of the particles, if the total energy of the particle is $E_{0}$

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$π \sqrt{\frac{m}{k}} + 4 \sqrt{\frac{2 E_{0}}{m g^{2}}}$

$π \sqrt{\frac{m}{k}} + 2 \sqrt{\frac{2 E_{0}}{m g^{2}}}$

$2 \sqrt{\frac{2 E_{0}}{m g^{2}}}$

$2 π \sqrt{\frac{m}{k}}$

answer is C.

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Detailed Solution

The left half of the graph is usual SHM.
$∴ t_{1} = π \sqrt{\frac{m}{k}}$
For the right part, $U = m g x$
At the origin, $U = m g x$
At the origin, $U = 0$
$∴ K = E_{0}$

As we move right the potential energy increases linearly with x. This is similar to a body that is thrown vertically upwards with initial velocity.
$∴ \frac{1}{2} m u^{2} = E_{0} \Rightarrow u = \sqrt{\frac{2 E_{0}}{m}}$
Time of flight till it reaches point of projection (i.e.origin) is
$t_{2} = \frac{2 u}{g} = \frac{2}{g} \sqrt{\frac{2 E_{0}}{m}} = 2 \sqrt{\frac{2 E_{0}}{m g^{2}}}$
Total time of oscillation is $T_{1} = t_{1} + t_{2} = π \sqrt{\frac{m}{k}} + 2 \sqrt{\frac{2 E_{0}}{m g^{2}}}$