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(A) Given: Mass $m_1 = 0.5 \ kg$,extension $x = 0.2 \ m$,$g = 10 \ m \ s^{-2}$.From Hooke's Law,$mg = kx$,so the spring constant $k = \frac{mg}{x} = \

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$0.628$

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(A) Given: Mass $m_1 = 0.5 \ kg$,extension $x = 0.2 \ m$,$g = 10 \ m \ s^{-2}$.From Hooke's Law,$mg = kx$,so the spring constant $k = \frac{mg}{x} = \frac{0.5 	imes 10}{0.2} = \frac{5}{0.2} = 25 \ N/m$.The time period of a spring-mass system is given by $T = 2\pi \sqrt{\frac{m}{k}}$.When the mass is replaced by $m_2 = 0.25 \ kg$,the new time period is $T = 2\pi \sqrt{\frac{0.25}{25}} = 2\pi \sqrt{0.01} = 2\pi 	imes 0.1 = 0.2\pi \ s$.Using $\pi \approx 3.14$,$T = 0.2 	imes 3.14 = 0.628 \ s$.

$A$ spring is stretched by $0.2 \ m$ when a mass of $0.5 \ kg$ is suspended to it. The time period of the spring when the $0.5 \ kg$ mass is replaced with a mass of $0.25 \ kg$ is (Acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $s$)

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