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A mass is suspended from the end of a spring. When the system is oscillating the amplitude of oscillation is 4cm and maximum kinetic energy of oscillation of the system is 1J. Then the force constant of the spring is

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2500 N/m

1250 N/m

500 N/m

250 N/m

answer is B.

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Detailed Solution

$\begin{array}{l} A = 4 \times 10^{- 2} m; TE = 1 J \\ TE = \frac{1}{2} {mω}^{2} A^{2} = \frac{1}{2} {KA}^{2} \\ \Rightarrow 1 = \frac{1}{2} \times K \times {(4 \times 10^{- 2})}^{2} \\ K = 1250 {Nm}^{- 1} \end{array}$

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