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One end of a spring of force constant k is fixed to a vertical wall and the other to a block of mass m resting on a smooth horizontal surface. There is another wall at a distance $x_{0}$ from the black. The spring is then compressed by $2 x_{0}$ and released. The time taken to strike the wall is

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a

$\frac{1}{6} \sqrt{\frac{k}{m}}$

b

$\sqrt{\frac{k}{m}}$

c

$\frac{2 π}{3} \sqrt{\frac{m}{k}}$

d

$\frac{π}{4} \sqrt{\frac{k}{m}}$

answer is C.

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Detailed Solution

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The total time from A to C
$t_{AC} = t_{AB} + t_{BC} = (T / 4) + t_{BC}$
where T = time period of oscillation of spring
mass system
t_BC can be obtained from,
BC = ABsin(2π / T)t_BC
Putting $\frac{BC}{AB} = \frac{1}{2}$ we obtain $t_{BC} = \frac{T}{12}$
$\Rightarrow t_{AC} = \frac{T}{4} + \frac{T}{12} = \frac{2 π}{3} \sqrt{\frac{m}{k}}$

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