Question #326ea

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Answer 1 · 2016-07-20T05:25:26

#3/4#

If a particle of mass m executing SHM with amplitude a and angular frequency w,then its KE at displacement x is

#KE-> E_k=1/2*mw^2(a^2-x^2)#

#"Total Energy"->E=1/2mw^2a^2#

KE when displacement is half the amplitude# x=a/2#

#E_(1/2)=1/2mw^2(a^2-a^2/4)#

#=1/2mw^2*3/4a^2=3/4E#

So the ratio is

#E_(1/2)/E=3/4=3/4#