Question

A particle undergoing simple harmonic motion has time dependent displacement given by $x(t) = A \sin \frac{\pi t}{90}$. The ratio of kinetic to potential energy of this particle at $t\, =\, 210\, s$ will be :

• A. 2
• B. $\frac{1}{9}$
• C. 3
• D. 1

Answer 1

Answer: (C)

3

Answer 2

$k = \frac{1}{2} m\omega^{2}A^{2} \cos^{2} \omega t$
$U = \frac{1}{2}m\omega^{2} A^{2} \sin^{2}\omega t$
$\frac{k}{U} =\cot^{2} \omega t =\cot^{2} \frac{\pi}{90} \left(210\right) =\frac{1}{3}$
Hence ratio is $3$ (most appropriate)