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If \( 10 cm \) is the amplitude of oscillation performed by a particle in SHM, then what fraction of total oscillation energy is kinetic when the particle is at \( 5 cm \) from the mean position? |
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Answer» Etotal \(= \frac{1}{2} m \omega^2 A^2\) Etotal \(= \frac{1}{2} \times m \omega^2 (A^2)\) Etotal \(= \frac{1}{2} \times m \omega^2 \times (10^2)\) Etotal \(= \frac{1}{2} \times m \omega^2 \times 100 ....(1)\) K.E. \(= \frac{1}{2} m \omega^2 (A^2 - x^2)\) K.E. \(= \frac{1}{2} m \omega^2 [(10)^2 - (5)^2]\) K.E. \(= \frac{1}{2} m \omega^2 [100 - 25]\) K.E. \(= \frac{1}{2} m \omega^2 \times 75 ....(2)\) Fraction of total energy and K.E. energy \(\frac{(1)}{(2)}\) \(\frac{E_{total}}{K.E.} = \cfrac{\frac{1}{2}m\omega^2 \times 100}{\frac{1}{2} m \omega^2 \times 75}\) \(\frac{E_{total}}{K.E.} = \frac{100}{75} = \frac{4}{3}\) \(\frac{E_{total}}{K.E.} = \frac{4}{3}.\) |
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