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Position of a $3 \mathrm{~kg}$ mass moving along the $X$-axis is given by $x=0.3 \cos (\omega t) \mathrm{m}$. If $K(t)$ denotes the kinetic energy at time $t$.
Then the value of $\frac{K\left(\frac{\pi}{6 \omega}\right)}{K\left(\frac{\pi}{3 \omega}\right)}$ is
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Then the value of $\frac{K\left(\frac{\pi}{6 \omega}\right)}{K\left(\frac{\pi}{3 \omega}\right)}$ is
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Verified Answer
The correct answer is:
$1 / 3$
Given, position of particle of mass $(m=3 \mathrm{~kg})$ is $x=0.3 \cos \omega t$
Velocity of particle, $v=\frac{d x}{d t}$
$\Rightarrow \quad v=\frac{d}{d t} 0.3 \cos \omega t$
$\begin{aligned} & =0.3 \frac{d}{d t} \cos \omega t=0.3(-\sin \omega t) \times \omega \\ & =-0.3 \omega \sin \omega t\end{aligned}$
Now, at $t=\frac{\pi}{6 \omega}$,
Velocity, $v_1=-0.3 \omega \sin \left(\omega \times \frac{\pi}{6 \omega}\right)$
$\Rightarrow \quad v_1=-0.3 \omega \cdot \sin \frac{\pi}{6}=\frac{-0.3 \omega}{2}$
At $\quad t=\frac{\pi}{3 \omega}$,
Velocity, $v_2=-0.3 \omega \sin \frac{\pi}{3}$
$v_2=-\frac{-0.3 \sqrt{3} \omega}{3}$
Ratio of kinetic energies,
$\begin{aligned} & \frac{K\left(\frac{\pi}{6 \omega}\right)}{K\left(\frac{\pi}{3 \omega}\right)}=\frac{\frac{1}{2} m v_1^2}{\frac{1}{2} m v_2^2}=\left(\frac{v_1}{v_2}\right)^2 \\ & \frac{K\left(\frac{\pi}{6 \omega}\right)}{K\left(\frac{\pi}{3 \omega}\right)}=\frac{\left(-\frac{0.3 \omega}{2}\right)^2}{\left(-0.3 \omega \cdot \frac{\sqrt{3}}{2}\right)^2}=\frac{1}{3}\end{aligned}$
Velocity of particle, $v=\frac{d x}{d t}$
$\Rightarrow \quad v=\frac{d}{d t} 0.3 \cos \omega t$
$\begin{aligned} & =0.3 \frac{d}{d t} \cos \omega t=0.3(-\sin \omega t) \times \omega \\ & =-0.3 \omega \sin \omega t\end{aligned}$
Now, at $t=\frac{\pi}{6 \omega}$,
Velocity, $v_1=-0.3 \omega \sin \left(\omega \times \frac{\pi}{6 \omega}\right)$
$\Rightarrow \quad v_1=-0.3 \omega \cdot \sin \frac{\pi}{6}=\frac{-0.3 \omega}{2}$
At $\quad t=\frac{\pi}{3 \omega}$,
Velocity, $v_2=-0.3 \omega \sin \frac{\pi}{3}$
$v_2=-\frac{-0.3 \sqrt{3} \omega}{3}$
Ratio of kinetic energies,
$\begin{aligned} & \frac{K\left(\frac{\pi}{6 \omega}\right)}{K\left(\frac{\pi}{3 \omega}\right)}=\frac{\frac{1}{2} m v_1^2}{\frac{1}{2} m v_2^2}=\left(\frac{v_1}{v_2}\right)^2 \\ & \frac{K\left(\frac{\pi}{6 \omega}\right)}{K\left(\frac{\pi}{3 \omega}\right)}=\frac{\left(-\frac{0.3 \omega}{2}\right)^2}{\left(-0.3 \omega \cdot \frac{\sqrt{3}}{2}\right)^2}=\frac{1}{3}\end{aligned}$
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