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  •  A Particle undergoing simple harmonic motion hal time dependent displacement given by <img alt="\small y(t)=A \cos \left(\frac{\bar{n} t}{6}\right)" src="https://entrancecorner.oncodecogs.com

 A Particle undergoing simple harmonic motion hal time dependent displacement given by \small y(t)=A \cos \left(\frac{\bar{n} t}{6}\right). The ratio of Kinetic to Potential energy of this Particle at \small t=360^{\circ} \mathrm{sec}

Option: 1

0


Option: 2

1


Option: 3

2


Option: 4

3


Answers (1)

best_answer

y=A \cos \left(\frac{\pi t}{60}\right)

\frac{\pi}{60}=\omega

V=-\frac{A \pi}{60} \sin \left(\frac{ \pi\ t}{60}\right)

P E=\frac{1}{2} m \omega^2 y^2=\frac{1}{2} m \omega^2 A^2 \cos ^2\left(\frac{\pi\ t}{60}\right)

K E=\frac{1}{2} m\left(-\frac{A \pi}{60}\right)^2 \sin ^2\left(\frac{\pi t}{60}\right)

K E=\frac{1}{2} m A^2 \frac{\pi^2}{60^2} \sin ^2\left(\frac{\pi t}{60}\right) \quad\left[\because \frac{\pi^2}{60^2}=\omega^2\right\].

\begin{aligned} & t=360 \\ & \frac{K E}{P E}=? \end{aligned}

\frac{K E}{P E}=\frac{\frac{1}{2} m A^2 \omega^2 \sin ^2\left(\frac{\pi t}{60}\right)}{\frac{1}{2} m \omega^2 A^2\left(\operatorname{cos}^2\left(\frac{\pi t}{60}\right)\right.}

\frac{K E}{P E}=\tan ^2\left(\frac{\pi t}{60}\right)=\tan ^2\left(\frac{\pi \times 360}{60}\right)

=\tan ^2(6 \pi)=\tan ^2(1080)

\frac{K E}{P E}=0

 

 

 

Posted by

Gautam harsolia

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