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A body of mass 25 gm performs linear S.H.M. The force constant of the motion is 400 dyne/cm. When the body is at a distance of 10 cm from the equilibrium position has velocity of 40 cm/s, then the total energy of the body will beA. `40xx10^(4)J`B. `4xx10^(4)erg`C. `2xx10^(4)erg`D. `2xx10^(5)J` |
Answer» Correct Answer - B `omega=sqrt((k)/(m))=sqrt((400)/(25))=(20)/(5)=4` `v=omegasqrt(A^(2)-x^(2))therefore 40=4sqrt(A^(2)-100)` `100=A^(2)-100 therefore A^(2)=200` `T.E.=(1)/(2)kA^(2)=(1)/(2)xx400xx200=4xx10^(4)erg` |
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