The ratio of the kinetic energy $E_{k}$ and potential energy of a particle executing SHM, at a distance x from mean position will be

\frac{x^{2}}{x^{2} - a^{2}}

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\frac{x^{2} - a^{2}}{x^{2}}

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\frac{x^{2}}{a^{2} - x^{2}}

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\frac{a^{2} - x^{2}}{x^{2}}

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Solution

The correct option is D $\frac{a^{2} - x^{2}}{x^{2}}$

$\frac{K E}{P E} = \frac{0.5 m v^{2}}{0.5 k x^{2}}$
$= \frac{m a^{2} ω^{2} c o s^{2} ω t}{m ω^{2} . a^{2} . s i n^{2} ω t} = \frac{c o s^{2} ω t}{s i n^{2} ω t}$
$= \frac{1 - s i n^{2} ω t}{s i n^{2} ω t}$
Now, $s i n ω t = x / a$
Thus, $\frac{K E}{P E} = \frac{1 - x^{2} / a^{2}}{x^{2} / a^{2}} = \frac{a^{2} - x^{2}}{x^{2}}$

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