On a frictionless horizontal plane, a bob of mass $m = 0.1 k g$ is attached to a spring with natural length $l_{0} = 0.1 m$ . The spring constant is $k_{1} = 0.009 N m^{- 1}$ when the length of the spring $l > l_{0}$ and is $k_{2} = 0.016 N m^{- 1}$ when $l < l_{0}$ . Initially the bob is released from $l = 0.15 m$ . Assume that Hooke’s law remains valid throughout the motion. If the time period of the full oscillation is $T = (n π) s$ , then the integer closest to n is ______

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answer is 6.

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Detailed Solution

For the given situation described in question the force constant of spring is different in case of compression or elongation state given as $l > l_{0} \to k = k_{1}$

$l < l_{0} \to k = k_{2}$

So for the two half of oscillations on the two sides of mean position, total time period of oscillation is given as $T = π \sqrt{\frac{m}{k_{1}}} + π \sqrt{\frac{m}{k_{2}}}$

$\Rightarrow T = π \sqrt{\frac{0.1}{0.009}} + π \sqrt{\frac{0.1}{0.016}}$

$\Rightarrow T = \frac{π}{0.3} + \frac{π}{0.4}$

$\Rightarrow T = \frac{0.7}{0.12} π \Rightarrow T = 5.83 π$

$\Rightarrow T = 6 π$

$\Rightarrow n = 6$

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