The variation of potential energy U of a harm | vedclass

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(B) From the given figure,the potential energy $U$ is given by $U = U_0 + \frac{1}{2}Ky^2$,where $U_0$ is the minimum potential energy at $y = 0$.At $

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$150 	ext{ Nm}^{-1}$

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(B) From the given figure,the potential energy $U$ is given by $U = U_0 + \frac{1}{2}Ky^2$,where $U_0$ is the minimum potential energy at $y = 0$.At $y = 0$,$U_{\min} = 0.01 	ext{ J}$.At $y = 20 	ext{ mm} = 20 	imes 10^{-3} 	ext{ m} = 2 	imes 10^{-2} 	ext{ m}$,$U_{\max} = 0.04 	ext{ J}$.The change in potential energy is $\Delta U = U_{\max} - U_{\min} = \frac{1}{2}Ky^2$.Substituting the values:$0.04 	ext{ J} - 0.01 	ext{ J} = \frac{1}{2} 	imes K 	imes (2 	imes 10^{-2} 	ext{ m})^2$$0.03 	ext{ J} = \frac{1}{2} 	imes K 	imes 4 	imes 10^{-4} 	ext{ m}^2$$0.03 = K 	imes 2 	imes 10^{-4}$$K = \frac{0.03}{2 	imes 10^{-4}} = \frac{3 	imes 10^{-2}}{2 	imes 10^{-4}} = 1.5 	imes 10^2 	ext{ Nm}^{-1} = 150 	ext{ Nm}^{-1}$.

The variation of potential energy $U$ of a harmonic oscillator with respect to displacement $y$ is as shown in the figure. Find the spring constant $K$.

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