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Q.

The mean kinetic energy of a particle executing SHM over one full oscillation, given  mass of particle = m, amplitude = A, angular frequency = ω ).

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a

14mω2A2

b

12mω2A2

c

122mω2A2

d

18mω2A2

answer is A.

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Detailed Solution

x=Asin⁡ωt So,v=Aωcos⁡ωtKE=12mv2=12m(Aωcos⁡ωt)2=12mω2A2cos2⁡ωtMean kinetic energy = KEmean =∫0T (KE)dt∫∫T dt=∫0T 12mω2A2cos2⁡ωtdtT=12mω2A2TT∫0T cos2⁡ωtdt=12mω2A2(T)∫0T 1+cos⁡2ωt2dt=12mω2A2Tt2+sin⁡2ωt4ω0T=12mω2A2TT2=14mω2A2.
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