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  • Find the displacement of a simple harmonic oscillator at which its PE is half of the maximum energy of the oscillator.

Find the displacement of a simple harmonic oscillator at which its PE is half of the maximum energy of the oscillator.

Answers (1)

Let us consider an oscillator viz. at ,

displacement = x from its mean position,

Mass = m

P.E. = \frac{1}{2} kx^{2}

Force constant of oscillator, k = m\omega ^{2}

Thus, P.E. = -m^{2}x^{2}

At x = A, when K.E. = 0, P.E. will be maximum & it will be the total energy of the oscillator

E = \frac{1}{2} m\omega ^{2}A^{2}

At displacement ‘x’, P.E. = \frac{1}{2}E

Thus,

\frac{1}{2} m\omega ^{2}x^{2}= \frac{1}{2}.\frac{1}{2} m\omega ^{2}A^{2}

& x = \pm \frac{A}{\sqrt{2}}

Thus, when displacement = \pm \frac{1}{\sqrt{2}} amplitude from mean position, P.E. will be half of the total energy.

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