A particle of mass 0.1 kg is executing simple harmonic motion with amplitude 20 cm along a straight line. Its maximum and minimum potential energies are 0.8 J and 0.6 J respectively. Then period of oscillation of the particle is

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π2 S

π4 S

π10 S

π5 S

answer is D.

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Detailed Solution

(4)Energy of oscillation E=Umax−Umin=0.8−0.6J=0.2J∴ 12 mω2A2=E ⇒ ω=1A2Em∴ T=2πω=2π.Am2E =2π × 0.2 0.12 × 0.2 sec=π5sec.

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