A linear harmonic oscillator of force constant $2\times {{10}^{6}}N{{m}^{-1}}$ and amplitude $0.01m$ has a total mechanical energy of $160J$. Its
$\left( \text{This question has multiple correct options} \right)$
A. Maximum potential energy is $100J$
B. Maximum kinetic energy is $100J$
C. Maximum potential energy is $160J$
D. Maximum potential energy is zero

Hint: For a linear harmonic oscillator the total mechanical energy is the sum of the potential energy and the kinetic energy of the oscillator. At the maximum displacement the kinetic energy becomes zero so that the total mechanical energy becomes the maximum potential energy and at the equilibrium or at mean position the potential energy becomes zero so that the total mechanical energy becomes the maximum kinetic energy of the given linear harmonic oscillator. Using this information we can choose the correct answers from the above given options.

Complete step by step answer:
Given data is
The force constant of the harmonic oscillator is $K=2\times {{10}^{6}}N{{m}^{-1}}$
Amplitude of the harmonic oscillator is $x=0.01m$
The total mechanical energy is given by $E=160J$
We know that at the maximum displacement the kinetic energy becomes zero.
The total mechanical energy becomes the maximum potential energy
Since given the total mechanical energy $E=160J$
The maximum potential energy at the maximum displacement is $P{}_{\max }=160J$
Hence option (C) is correct.
We know that the maximum kinetic energy is
$KE=\dfrac{1}{2}K{{\left( {{x}_{\max }} \right)}^{2}}$
Substitute the above given values in this formula
$\Rightarrow KE=\dfrac{1}{2}\left( 2\times {{10}^{6}} \right){{\left( 0.01 \right)}^{2}}$
$\Rightarrow KE=100J$
Therefore the maximum kinetic energy is $KE=100J$
Hence option (B) is also correct.
Therefore both the options (B) & (C) are correct.

Note:
The total energy of the harmonic oscillator is the sum of the kinetic and the potential energy i.e., both will contribute at every position besides at the mean and the maximum position.