The energy of a particle executing simple harmonic motion is given by $E = Ax^2 + Bv^2$, where $x$ is the displacement from mean position $x = 0$ and $v$ is the velocity of the particle at $x$ then choose the incorrect statement
Diffcult
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$\mathrm{E}=\mathrm{Ax}^{2}+\mathrm{Bv}^{2}$
Velocity is maximum, when $\mathrm{x}=0 ; \mathrm{V}_{\max }=\sqrt{\frac{\mathrm{E}}{B}}$
$\{\text { Also } A=\mathrm{K} / 2, \mathrm{B}=\mathrm{m} / 2\}$
So $\omega=\sqrt{\frac{\mathrm{K}}{\mathrm{m}}}=\sqrt{\frac{\mathrm{A}}{\mathrm{B}}}$
when $\mathrm{v}=0, \mathrm{x}_{\max }=$ amplitude
$\mathrm{x}_{\max }=\sqrt{\frac{\mathrm{E}}{\mathrm{A}}}$
Time period $=\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{\mathrm{A} / \mathrm{B}}}=2 \pi \sqrt{\frac{\mathrm{B}}{\mathrm{A}}}$
$\max .$ acceleation $\omega^{2} \mathrm{x}_{\max }=\frac{\mathrm{A}}{\mathrm{B}} \sqrt{\frac{\mathrm{E}}{\mathrm{A}}}=\frac{\sqrt{\mathrm{EA}}}{\mathrm{B}}$
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