Question

Please answer this ​

Answer 1

Answer:

Given:

v1 and v2 are the velocities at x1 and x2 from mean position.

To find:

Angular frequency of the particle

Calculation:

For 1st case :

$v1 = \omega \sqrt{ { a}^{2} - {(x1)}^{2} }$

$= > {(v1)}^{2} = { (\omega a)}^{2} - {( \omega \times x1)}^{2}$

$= > {( \omega a)}^{2} = {(v1)}^{2} + {( \omega \times x1)}^{2}$

For 2nd case

$v2 = \omega \sqrt{ { a}^{2} - {(x2)}^{2} }$

$= > {( \omega a)}^{2} = {(v2)}^{2} + {( \omega \times x2)}^{2}$

Comparing (ωa)² :

$= > {(v2)}^{2} + {( \omega \times x2)}^{2} = {(v1)}^{2} + {( \omega \times x1)}^{2}$

$= > { \omega}^{2} ( {x2}^{2} - {x1}^{2} ) = {(v1)}^{2} - {(v2)}^{2}$

$= > { \omega}^{2} = \dfrac{ {(v1)}^{2} - {(v2)}^{2} }{ {(x2)}^{2} - {(x1)}^{2} }$

Taking minus sign common :

$= > { \omega}^{2} = \dfrac{ {(v2)}^{2} - {(v1)}^{2} }{ {(x1)}^{2} - {(x2)}^{2} }$

$= > { \omega} = \sqrt \dfrac{ {(v2)}^{2} - {(v1)}^{2} }{ {(x1)}^{2} - {(x2)}^{2} }$

So final answer is :

$\boxed{ \red{ { \omega} = \sqrt \dfrac{ {(v2)}^{2} - {(v1)}^{2} }{ {(x1)}^{2} - {(x2)}^{2} } }}$