Question

A particle is performing SHM in a straight line. If the velocities of the particle are $v_1$ and $v_2$ at displacements $x_1$ and $x_2$ respectively, the angular frequency of the particle is

• A. $\sqrt{\frac{v_1^2+v_2^2}{x_2^2+x_1^2}}$
• B. $\sqrt{\frac{v_1^2-v_2^2}{x_2^2+x_1^2}}$
• C. $\sqrt{\frac{v_1^2-v_2^2}{x_1^2-x_2^2}}$
• D. $\sqrt{\frac{v_1^2-v_2^2}{x_2^2-x_1^2}}$

Answer 1

The correct option is D $\sqrt{\frac{v_1^2-v_2^2}{x_2^2-x_1^2}}$

$v=\omega \sqrt{A^2-x^2}$
$\therefore v_1=\omega \sqrt{A^2-x_1^2}$ and $v_2\omega \sqrt{A^2-x_2^2}$
or $v_1^2={\omega }^2(A^2-x_1^2)$ and $v_2^2={\omega }^2(A^2-x_2^2)$
$\therefore v_1^2-v_2^2=(x_2^2-x_1^2){\omega }^2\Rightarrow {\omega }^2=\frac{v_1^2-v_2^2}{x_2^2-x_1^2}$
$\Rightarrow \omega =\sqrt{\frac{v_1^2-v_2^2}{x_2^2-x_1^2}}$