Q132 / 160 · AP EAMCET 2024 (18 May Shift 1) · Physics
For a particle executing simple harmonic motion, match the following statements (conditions) from column I to statements (shapes of graph) in column II
- A(A) - (iv), (B) - (i), (C) - (ii), (D) - (iii)
- Correct answerB(A) - (iii), (B) - (i), (C) - (ii), (D) - (iv)
- C(A) - (iii), (B) - (ii), (C) - (i), (D) - (iv)
- D(A) - (iv), (B) - (ii), (C) - (i), (D) - (iii)
Solution
For a particle executing SHM, $\mathrm{x}=\mathrm{A} \sin \omega \mathrm{t}, \mathrm{v}=\mathrm{AW} \cos \omega \mathrm{t}, \mathrm{a}=-\mathrm{A}\omega^2 \sin \omega \mathrm{t}$
$\therefore v=A \omega \sqrt{1-\sin ^2 \omega t}=A \omega \sqrt{1-\left(\frac{x}{A}\right)^2}=\omega \sqrt{A^2-x^2}$
At $\omega=1, v^2=A^2-x^2 \Rightarrow v^2+x^2=A^2$
$\therefore$ Velocity displacement graph is a circle.
Now, $a=-A \omega^2 \sin \omega t$
$\therefore \quad$ Acceleration time graph is sinusoidal
Also, $a=\left(A \omega^2\right)\left(\frac{x}{A}\right)=-\omega^2 x$
$\therefore$ Acceleration displacement graph is straight line
Now, $v=A \omega \cos \omega t,-a=-A \omega^2 \sin \omega t$
$\therefore\left(\frac{\mathrm{v}}{\mathrm{A}\omega}\right)^2+\left(\frac{\mathrm{a}}{-\mathrm{A}\omega}\right)^2=\sin ^2 \omega \mathrm{t}+\cos ^2 \mathrm{wt}=1$
$\Rightarrow \frac{v^2}{A^2 \omega^2}+\frac{a^2}{A^2 \omega^4}=1$
$\therefore$ Acceleration - velocity graph is ellipse
Hence, the correct answer is option: 'B'.