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The variation of potential energy of a harmonic oscillator is as shown in the figure. Then, find the spring constant.
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Verified Answer
The correct answer is:
\(150 \mathrm{Nm}^{-1}\)
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,
\(\begin{array}{ll}
& U_{\max }-U_{\min }=\frac{1}{2} K y^2 \\
\Rightarrow & 4 \times 10^{-2}-1 \times 10^{-2}=\frac{1}{2} K \times\left(2 \times 10^{-2}\right)^2 \\
\Rightarrow & 3 \times 10^{-2}=K \times 2 \times 10^{-4} \\
\Rightarrow & \quad K=\frac{3 \times 10^{-2}}{2 \times 10^{-4}}=1.5 \times 10^2=150 \mathrm{Nm}^{-1}
\end{array}\)
\(=2 \times 10^{-2} \mathrm{~m}\)
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,
\(\begin{array}{ll}
& U_{\max }-U_{\min }=\frac{1}{2} K y^2 \\
\Rightarrow & 4 \times 10^{-2}-1 \times 10^{-2}=\frac{1}{2} K \times\left(2 \times 10^{-2}\right)^2 \\
\Rightarrow & 3 \times 10^{-2}=K \times 2 \times 10^{-4} \\
\Rightarrow & \quad K=\frac{3 \times 10^{-2}}{2 \times 10^{-4}}=1.5 \times 10^2=150 \mathrm{Nm}^{-1}
\end{array}\)
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