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A block of mass 0.2 kg performs simple harmonic motion along the X-axis with a frequency of (25/pi) Hz. At the position x = 0.04 m, its kinetic energy is 0.5 J and its potential energy is 0.4 J (taking equilibrium as zero PE). Find the amplitude of oscillation.

1. 2 cm
2. 4 cm
3. 5 cm
4. 6 cm

Correct answer: 6 cm

Solution

At any point in SHM the total energy E = KE + PE = 0.5 + 0.4 = 0.9 J. With omega = 2 pi f = 50 rad/s and E = (1/2)m omega² A², solve for A.

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