Oscillations

JEE Main level, numerical answer, medium

76 of 194

Question 76 of 194

When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is x8\dfrac{x}{8}, where x=x=_________.

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Answer: 9

The total energy of the harmonic oscillator is given by the formula
E=12kA2...(1)E=\dfrac{1}{2}k{A}^{2}...\left(1\right)
where, kk is the force constant and AA is the amplitude.
The potential energy of the oscillator at the given displacement can be written as
U=12k(A3)2=kA22×9=E9\begin{matrix}U & = & \dfrac{1}{2}k{\left(\dfrac{A}{3}\right)}^{2} \\ & = & \dfrac{k{A}^{2}}{2\times 9} \\ & = & \dfrac{E}{9}\end{matrix}
Hence, its kinetic energy is given by
KE=E−E9=8E9\begin{matrix}KE & = & E-\dfrac{E}{9} \\ & = & \dfrac{8E}{9}\end{matrix}
Thus, the required ratio is
EKE=E8E9=98\begin{matrix}\dfrac{E}{KE} & = & \dfrac{E}{\dfrac{8E}{9}} \\ & = & \dfrac{9}{8}\end{matrix}
Hence, x=9x=9.

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