Two simple harmonic motions $y_1 = A \sin \omega t$ and $y_2 =A \cos \omega t$ are superimposed on a particle of mass $m.$ The total mechanical energy of the particle is :
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The two $S H M^{\prime} s$ $y_{1}$ and $y_{2}$ are superimposed to give a new $S H M$ as $y .$
$y=y_{1}+y_{2}$
$y=A \sin (w t)+A \cos (w t)$
$y=\sqrt{2} A\left[\frac{1}{\sqrt{2}} \times \sin (w t)+\frac{1}{\sqrt{2}} \times \cos (w t)\right]$
$y=\sqrt{2} A\left[\cos \left(\frac{\pi}{4}\right) \sin (w t)+\sin \left(\frac{\pi}{4}\right) \cos (w t)\right] \quad \Longrightarrow \quad y=\sqrt{2} A[\sin (w t+\pi / 4)]$
But for $SHM$ $y=B \sin (w t+\phi), \quad$ total mechanical energy $\quad T \cdot E=\frac{1}{2} m w^{2} B^{2}$
Thus total mechanical energy for $y, \quad T . E=\frac{1}{2} m w^{2}(\sqrt{2} A)^{2}$
$\Longrightarrow \quad T . E=m w^{2} A^{2}$
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