Search any question & find its solution
Question:
Answered & Verified by Expert
A particle is moving in $x-y$ plane according to $\vec{r}=b \cos \omega t \hat{i}+b \sin \omega t \hat{j}$. Where $\omega$ is constant. Which of the following statement(s) is / are true?
Options:
Solution:
2643 Upvotes
Verified Answer
The correct answers are:
$\frac{E}{\omega}$ is a constant where $E$ is the total energy of the particle, The trajectory of the particle in $x-y$ plane is a circle
$\vec{r}=b \cos \omega t i+b \sin \omega t j$
$\vec{v}=\frac{d \vec{r}}{d t}=-b \omega \sin \omega t+b \omega \cos \omega t j$
$(\vec{v})=\sqrt{b^2 \omega^2 \sin ^2 \omega t+b^2 \omega^2 \cos ^2 \omega t}$
$(\vec{v})=b \omega \rightarrow$ constant $\quad \therefore E=\frac{1}{2} m v^2=$ constant
$E=\frac{1}{2} m\left[b^2 \omega^2\right] \quad \Rightarrow \frac{E}{\omega}=\frac{1}{2} m b^2 \cdot \omega=$ constant
$x=b \cos \omega t$
$y=b \sin \omega t$
$x^2+y^2=b^2 \rightarrow$ equation of circle
$\therefore \vec{a} \perp \vec{v} \quad$ So $\vec{a} \neq \omega^2 \vec{v}$
$\vec{v}=\frac{d \vec{r}}{d t}=-b \omega \sin \omega t+b \omega \cos \omega t j$
$(\vec{v})=\sqrt{b^2 \omega^2 \sin ^2 \omega t+b^2 \omega^2 \cos ^2 \omega t}$
$(\vec{v})=b \omega \rightarrow$ constant $\quad \therefore E=\frac{1}{2} m v^2=$ constant
$E=\frac{1}{2} m\left[b^2 \omega^2\right] \quad \Rightarrow \frac{E}{\omega}=\frac{1}{2} m b^2 \cdot \omega=$ constant
$x=b \cos \omega t$
$y=b \sin \omega t$
$x^2+y^2=b^2 \rightarrow$ equation of circle
$\therefore \vec{a} \perp \vec{v} \quad$ So $\vec{a} \neq \omega^2 \vec{v}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.