The variation of potential energy of a harmon | vedclass

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Question

(B) From the given graph,the potential energy $U$ varies with displacement $y$. The potential energy of a harmonic oscillator is given by $U(y) = U_0

Vedclass · Accepted Answer

$150 	ext{ N/m}$

Vedclass · Answer

(B) From the given graph,the potential energy $U$ varies with displacement $y$. The potential energy of a harmonic oscillator is given by $U(y) = U_0 + \frac{1}{2}ky^2$,where $U_0$ is the minimum potential energy at the equilibrium position $(y=0)$.From the graph,the minimum potential energy at $y=0$ is $U_0 = 0.01 	ext{ J}$.The maximum potential energy at the extreme position $y = 20 	ext{ mm} = 0.02 	ext{ m}$ is $U_{max} = 0.04 	ext{ J}$.Therefore,the change in potential energy from equilibrium to the extreme position is $\Delta U = U_{max} - U_0 = 0.04 	ext{ J} - 0.01 	ext{ J} = 0.03 	ext{ J}$.This change in potential energy is equal to the elastic potential energy stored in the spring,which is $\frac{1}{2}kA^2$,where $A$ is the amplitude.$0.03 = \frac{1}{2} 	imes k 	imes (0.02)^2$$0.03 = \frac{1}{2} 	imes k 	imes 0.0004$$0.03 = k 	imes 0.0002$$k = \frac{0.03}{0.0002} = \frac{300}{2} = 150 	ext{ N/m}$.

The variation of potential energy of a harmonic oscillator is as shown in the figure. The spring constant is

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