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Question

A particle executes SHM on a line 8 cm long. Its K.E. and P.E. will be equal when distance from the mean position is :-

A
4 cm
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B
2 cm
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C
22cm
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D
2cm
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Solution

The correct option is C 22cm
Let equation of SHM is x=Asin(ωt), WHere A is amplitude and ω is angular velocity.
v=Aωcos(ωt)
Kinetic Energy(KE)=12mv2=12mA2ω2cos2(ωt)=12mA2ω2(1sin2(ωt))
Kinetic Energy=12mω2(A2x2),
Total Energy = Kinetic Energy + Potential Energy
Total Energy at mean position= Kinetic Energy ( PE=0)
Total Energy=12mA2ω2
Let at position x Kinetic Energy and potential Energy will be equal.
KE=12mω2(A2x2)
Potential Energy= Total Energy - Kinetic Energy=12mA2ω212mω2(A2x2)=12mω2x2
KE=PE 12mω2(A2x2)=12mω2x2 x=A2
Given A=4cm x=22cm

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