An object of mass 0.2 kg executes simple harmonic motion along x axis with frequency of $$\left(\frac{25}{\pi}\right)$$ Hz. At the position $$x = 0.04$$ m the object has kinetic energy 0.5 J and potential energy 0.4 J. The amplitude of oscillation is _____ cm.

Correct Answer: 6

Total energy E = KE+PE = 0.5+0.4 = 0.9 J. E = ½mω²A². ω = 2πf = 2π×25/π = 50 rad/s. 0.9 = ½×0.2×2500×A². A² = 0.0036. A = 0.06 m = 6 cm.

The answer is 6.

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