Questions
A particle executes SHM of time period T. The time taken by the particle to move from mean position to the position where its kinetic energy is equal to potential energy is
a
T3
b
T4
c
T6
d
T8
detailed solution
Correct option is D
x=Asinωtv=ωA2−x2 So, KE=12mv2=12mω2A2−12mω2x2 And PE=12Kx2=12mω2x2PE=KE⇒12mω2A2−12mω2x2=12mω2x2⇒x=±A2 Let us consider motion from x=0 to x=A2A2=Asinωtωt=π4,3π4,9π4,11π4,…⇒t=T8,3T8,9T8,11T8,..Talk to our academic expert!
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When the displacement is half the amplitude, the ratio of potential energy to the total energy is
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