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A particle of mass ' $\mathrm{m}$ ' is executing simple harmonic motion about its mean position. If $' A^{\prime}$ is the amplitude and ' $T$ ' is the period of S.H.M., then the total energy of the particle is
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The correct answer is:
$\frac{2 \pi^{2} \mathrm{~mA}^{2}}{\mathrm{~T}^{2}}$
\(x=A \sin \omega t, \quad,=\frac{d x}{d t}=\omega A \cos \omega t\)
\(T \cdot E=P \cdot E+K \cdot E, P_{0} E=\frac{1}{2} k x^{2}, K \cdot E =\frac{1}{2} m v^{2}\)
T. \(E=\frac{1}{2} \times k \times A^{2} \sin ^{2} \omega t+\frac{1}{2} m \omega^{2} A^{2} \cos ^{2} \omega t \quad, \omega=\sqrt{\frac{R}{m}}, k mv^{2}\)
T. \(E=\frac{1}{2} m \omega^{2} A^{2} \sin ^{2} \omega t+\frac{1 m \omega^{2} A^{2} c c^{2} \omega t}{2} \Rightarrow \frac{m \omega^{2} A^{2}}{2}=\frac{2 m \pi^{2} A^{2}}{T^{2}}\)
\(T \cdot E=P \cdot E+K \cdot E, P_{0} E=\frac{1}{2} k x^{2}, K \cdot E =\frac{1}{2} m v^{2}\)
T. \(E=\frac{1}{2} \times k \times A^{2} \sin ^{2} \omega t+\frac{1}{2} m \omega^{2} A^{2} \cos ^{2} \omega t \quad, \omega=\sqrt{\frac{R}{m}}, k mv^{2}\)
T. \(E=\frac{1}{2} m \omega^{2} A^{2} \sin ^{2} \omega t+\frac{1 m \omega^{2} A^{2} c c^{2} \omega t}{2} \Rightarrow \frac{m \omega^{2} A^{2}}{2}=\frac{2 m \pi^{2} A^{2}}{T^{2}}\)
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