The particle executing simple harmonic motion has a kinetic energy K_(0)cos^(2) omega t. The maximum value of potential energy and the total energy are respectively…………and…………..

The particle executing simple harmonic motion has a kinetic energy K_(

1.	The particle executing simple harmonic motion has a kinetic energy K_(0)cos^(2) omega t. The maximum value of potential energy and the total energy are respectively…………and…………..
Answer» `(K_0)/(2)" and "K_(0)` `(K_0)" and "2K_(0)` `(K_0)" and "K_(0)` `0" and "2K_(0)` SOLUTION :Total energy of SHO E= U+K When KINETIC energy X is zero then its potential energy is maximum. Means `U_("max")+0=E`. Hence `U_("max")= E"""……."(1)` Kinetic energy `=(1)/(2) m OMEGA^(2) a^(2) cos^(2) omega t` `= K_(0) cos^(2) omega t` (is given) `therefore K_(0)= (1)/(2) m omega^(2) a^(2)=` Total mechanical energy Now, total mechanical energy `= K_(0)` and maximum potential energy `U_("max")= E= K_(0)`.