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Q.

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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a

π2 S

b

π4 S

c

π10 S

d

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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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