An object of mass $0.2\ kg$ executes simple harmonic along $X-$ axis with frequency of $\frac{{25}}{\pi } Hz$ . At the position $x$ = $0.04\ m$ , the object has kinetic energy of $0.5\ J$ and potential energy of $0.4\ J$ amplitude of oscillation in meter is equal to
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$\mathrm{E}=\frac{1}{2} \mathrm{m} \omega^{2} \mathrm{A}^{2} \Rightarrow \mathrm{E}=\frac{1}{2} \mathrm{m}(2 \pi \mathrm{f})^{2} \mathrm{A}^{2}$
$\Rightarrow \mathrm{A}=\frac{1}{2 \pi \mathrm{f}} \sqrt{\frac{2 \mathrm{E}}{\mathrm{m}}}$
Putting $\mathrm{E}=\mathrm{K}+\mathrm{U}$ we obtain,
$A=\frac{1}{2 \pi\left(\frac{25}{\pi}\right)} \sqrt{\frac{2 \times(0.5+0.4)}{0.2}} \Rightarrow A=0.06$
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