The angular velocity and amplitude of a simple pendulum are ‘ $ω$ ’ and A respectively. At a displacement ‘x’. If its KE is ‘T’ and PE is ‘U’ then the ratio of 'T' to ‘U’ is

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$\frac{x^{2} ω^{2}}{A^{2} - ω^{2} x^{2}}$

$\frac{x^{2}}{A^{2} - x^{2}}$

$\frac{A^{2} - ω^{2} x^{2}}{x^{2} ω^{2}}$

$\frac{A^{2} - x^{2}}{x^{2}}$

answer is D.

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Detailed Solution

$\begin{array}{l} K \cdot E = \frac{1}{2} {mω}^{2} (A^{2} - x^{2}) = T \\ P . E = \frac{1}{2} {mω}^{2} x^{2} = U \\ \frac{T}{U} = \frac{\frac{1}{2} {mω}^{2} (A^{2} - x^{2})}{\frac{1}{2} {mω}^{2} x^{2}} = \frac{A^{2} - x^{2}}{x^{2}} \end{array}$

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