Statement I: In a simple harmonic motion the kinetic and potential energies become equal when the displacement is $\frac{1}{\sqrt{2}}$ times the amplitude.
Statement II: In SHM kinetic energy is zero when potential energy is maximum.

Question

Infinity Learn · Accepted Answer

From the relation, potential energy of SHM is given as $$PE$$ $$=$$ $$12m$$ω$$2y2$$ and $$KE$$ $$=$$ $$12m$$ω$$2(A2-y2)where$$ y is the displacement and A is the amplitufde By the condition $$KE$$ $$=$$ $$PE12m$$ω$$2(A2-y2)$$ $$=$$ $$12m$$ω$$2y2$$⇒mω$$2y2$$ $$=$$ $$12m$$ω$$2A2$$⇒ y $$=$$ $$A2Also$$ we know that total energy is E $$=$$ $$KE$$ $$+$$ PEHence, if $$PE$$ is maximum then $$KE$$ will be zero.

Statement I: In a simple harmonic motion the kinetic and potential energies become equal when the displacement is $\frac{1}{\sqrt{2}}$ times the amplitude.
Statement II: In SHM kinetic energy is zero when potential energy is maximum.

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Statement I: In a simple harmonic motion the kinetic and potential energies become equal when the displacement is 12 times the amplitude. Statement II: In SHM kinetic energy is zero when potential energy is maximum.

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Statement I: In a simple harmonic motion the kinetic and potential energies become equal when the displacement is $\frac{1}{\sqrt{2}}$ times the amplitude.
Statement II: In SHM kinetic energy is zero when potential energy is maximum.