For a particle executing simple harmonic motion, the kinetic energy K is given by, $$K0cos2$$⁡ωt. The maximum value of $$PE$$ is

Question

For a particle executing simple harmonic motion,  the kinetic energy K is given by, $$K0cos2$$⁡ωt. The maximum value of $$PE$$ is

Infinity Learn · Accepted Answer

$$KE$$ $$=K0$$ $$cos2$$ωt we know $$KE$$$$=12m(A$$ω $$cos$$ω$$t)2$$ ; if displacement $$y=Asin$$ωt maximum $$KE$$ $$=K0=12m(A$$ω$$)2$$ $$PE$$$$=12m$$ω$$2(Asin$$ω$$t)2$$ maximum $$PE$$$$=12m(A$$ω$$)2=K0$$ for maximum value of $$PE$$ or $$KE$$ , $$sin$$ω$$t=1=$$$$cos$$ωtThe mechanical energy of a simple harmonic motion is conserved. Since K $$=$$ $$K0$$  $$cos2$$ωt, the maximum value of Kinetic energy $$=$$ $$K0$$ $$=$$ maximum value of potential  energy at the extreme position of the body in simple harmonic motion.

For a particle executing simple harmonic motion, the kinetic energy K is given by, $K_{0} \cos^{2} ω t$ . The maximum value of PE is

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For a particle executing simple harmonic motion, the kinetic energy K is given by, K0cos2⁡ωt. The maximum value of PE is

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For a particle executing simple harmonic motion, the kinetic energy K is given by, $K_{0} \cos^{2} ω t$ . The maximum value of PE is