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A simple harmonic oscillator has amplitude 'A'. If the kinetic energy of oscillator isone-fourth of the total energy, then thedisplacement is(A) AV(B) V3 A/2(C) A/4(D) A/2 |
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Answer» :Amplitude of motion = AWE have to find displacement of simple harmonic oscillator from MEAN POSITION at where kinetic energy of oscillator becomes one-fourth of the total energy★ P.E. at displacement x from the mean position,U = 1/2 kx²★ K.E. at displacement x from the mean position,K = 1/2 k (A² - x²)★ Total energy at any pointE = K + U = 1/2 kA²where A denotes amplitudeLet p be the point at where kinetic energy becomes one-fourth of the total energy.➠ E/4 = K➠ 1/2 kA² = 4 × [1/2 k (A² - x²)]➠ A² = 4A² - 4x²➠ 4x² = 3A²➠ x² = 3A²/4➠ x = √3A/2∴ (B) is the correct answer. |
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