A spring is stretched by $5 ext{ cm}$ by a force $10 ext{ N}$. The time period of the oscillations when a

Question

A spring is stretched by $5 	ext{ cm}$ by a force $10 	ext{ N}$. The time period of the oscillations when a mass of $2 	ext{ kg}$ is suspended by it is:

Accepted Answer

1.  **Calculate Spring Constant ($k$):** According to Hooke's Law, the magnitude of the restoring force $F$ is related to the extension $x$ by $F = kx$ .
    *   Given Force ($F$) = $10 	ext{ N}$
    *   Extension ($x$) = $5 	ext{ cm} = 0.05 	ext{ m}$
    *   $k = \frac{F}{x} = \frac{10}{0.05} = 200 	ext{ N/m}$
2.  **Calculate Time Period ($T$):** The time period of a spring-mass system is given by the formula $T = 2\pi \sqrt{\frac{m}{k}}$ .
    *   Given Mass ($m$) = $2 	ext{ kg}$
    *   $T = 2\pi \sqrt{\frac{2}{200}} = 2\pi \sqrt{\frac{1}{100}}$
    *   $T = 2\pi \left( \frac{1}{10} ight) = \frac{2 	imes 3.14}{10}$
    *   $T = \frac{6.28}{10} = 0.628 	ext{ s}$

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