A particle is executing simple harmonic motion with a time period of $$2$$ second. At t $$=$$ $$0$$, the particle was in its extreme position. Then the time after which the potential energy of the particle will be equal to $$50$$% of its total energy is

Question

Infinity Learn · Accepted Answer

If E be the total energy and V be the potential energy at time t, then $$V=E2(1+$$$$cos$$⁡$$2$$ω$$t)$$, where ω$$=$$ angular frequency of SHM.when $$V=E/2$$, we get, $$E2=E2(1+$$$$cos$$⁡$$2$$ω$$t)$$⇒$$2$$ω$$t=$$π$$2$$⇒$$t=$$π$$T42$$π$$=T8t=28sec$$⁡$$=0.25sec$$⁡

A particle is executing simple harmonic motion with a time period of 2 second. At t = 0, the particle was in its extreme position. Then the time after which the potential energy of the particle will be equal to 50% of its total energy is

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