A mass of 5, {kg} is connected to a spring. T | vedclass

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(C) From the potential energy curve,the maximum potential energy is $U_{\max} = 10\, {J}$ at an amplitude $A = 2\, {m}$ (since the equilibrium positio

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(C) From the potential energy curve,the maximum potential energy is $U_{\max} = 10\, {J}$ at an amplitude $A = 2\, {m}$ (since the equilibrium position is at $x = 2\, {m}$ and the maximum displacement is $4\, {m} - 2\, {m} = 2\, {m}$).Using the formula $U_{\max} = \frac{1}{2} k A^2$,we have $10 = \frac{1}{2} k (2)^2$.This simplifies to $10 = 2k$,so the spring constant $k = 5\, {N/m}$.The time period of the spring-mass system is $T_{	ext{spring}} = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{5}{5}} = 2\pi\, {s}$.The time period of a simple pendulum is $T_{	ext{pendulum}} = 2\pi \sqrt{\frac{L}{g}}$.Given $T_{	ext{spring}} = T_{	ext{pendulum}}$,we equate $2\pi \sqrt{\frac{5}{5}} = 2\pi \sqrt{\frac{4}{g}}$.Squaring both sides,we get $1 = \frac{4}{g}$,which gives $g = 4\, {m/s^2}$.

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