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

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Aicharya · Accepted Answer

According to figure, when $y=20 \mathrm{~mm}$ $=2 \times 10^{-2} \mathrm{~m}$ then $U_{\max }=0.04 \mathrm{~J}=4 \times 10^{-2} \mathrm{~J}$ When $y=0$ then $U_{\min }=0.01 \mathrm{~J}=1 \times 10^{-2} \mathrm{~J}$ $\therefore$ The change in potential energy, $\begin{array}{ll} & U_{\max }-U_{\min }=\frac{1}{2} K y^2 \ \Rightarrow & 4 \times 10^{-2}-1 \times 10^{-2}=\frac{1}{2} K \times\left(2 \times 10^{-2}\right)^2 \ \Rightarrow & 3 \times 10^{-2}=K \times 2 \times 10^{-4} \ \Rightarrow & \quad K=\frac{3 \times 10^{-2}}{2 \times 10^{-4}}=1.5 \times 10^2=150 \mathrm{Nm}^{-1} \end{array}$

The variation of potential energy of a harmonic oscillator is as shown in the figure. Then, find the spring…

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