An object of mass $0.2\, kg$ executes simple harmonic along $X-$ axis with frequency of $\frac{{25}}{\pi }Hz$. At the position $x = 0.04m$, 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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(b) $E = \frac{1}{2}m{\omega ^2}{A^2}$$ \Rightarrow $$E = \frac{1}{2}m\,{(2\pi f)^2}{A^2}$$ \Rightarrow $$A = \frac{1}{{2\pi f}}\sqrt {\frac{{2E}}{m}} $
Putting $E = K + U$ 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\,m$
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