An object of mass $0.2 \mathrm{~kg}$ executes simple harmonic motion along $\mathrm{x}$ axis with frequency of $\left(\frac{25}{\pi}\right) \mathrm{Hz}$. At the position $\mathrm{x}=0.04 \mathrm{~m}$ the object has kinetic energy $0.5 \mathrm{~J}$ and potential energy $0.4 \mathrm{~J}$ The amplitude of oscillation is ............ cm.
JEE MAIN 2024, Diffcult
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$\text { Total energy }=\text { K.E. }+ \text { P.E. }$
$\text { at } \mathrm{x}=0.04 \mathrm{~m}, \text { T.E. }=0.5+0.4=0.9 \mathrm{~J}$
$\text { T.E }=1 \mathrm{~m}^2 \mathrm{~A}^2=0.9$
$\quad=\frac{1}{2} \times 0.2\left(2 \pi \times \frac{25}{\pi}\right)^2 \times \mathrm{A}^2=0.9$
$\Rightarrow \mathrm{A}=0.06 \mathrm{~m}$
$\mathrm{~A}=6 \mathrm{~cm}$
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