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. $ \frac{1}{3}$
• D. 1

Answer 1

Answer: (C)

Kinetic Energy, $K =\frac{1}{2} m \omega^{2} A ^{2} \cos ^{2} \omega t$
Potential Energy, $U=\frac{1}{2} m \omega^{2} A^{2} \sin ^{2} \omega t$
$
\frac{ K }{ U }=\cot ^{2} \omega t =\cot ^{2} \frac{\pi}{90}(210)=\cot ^{2}\left(2 \pi+\frac{\pi}{3}\right)=\frac{1}{3}
$
Hence ratio is $\frac{1}{3}$.