Question

in simple harmonic motion, velocity is alpha and acceleration is bita, so A= ? and the answer is √α^2y-βy^2/β​

Answer 1

Answer:

An object is performing SHM , with instantaneous Acceleration β and instantaneous velocity being α

Now , we need to find a relationship between Amplitude , Acceleration and velocity .

$acc. = - { \omega}^{2} y = \beta$

$velocity = \omega \sqrt{ {A}^{2} - {y}^{2} } = \alpha$

Putting the value of Velocity in terms of angular frequency in the acceleration equation :

$\beta = - \bigg \{\dfrac{ { \alpha }^{2} }{ {A}^{2} - {y}^{2} } \bigg \} y$

Transposing to other sides :

$= > {A}^{2} - {y}^{2} = - \dfrac{ { \alpha }^{2} y}{ \beta }$

$= > {A}^{2} = {y}^{2} - \dfrac{ { \alpha }^{2}y }{ \beta }$

$= > A = \sqrt{ {y}^{2} - \dfrac{ { \alpha }^{2}y }{ \beta } }$

So final answer :

$\boxed{ \red{ \huge{ \bold{A = \sqrt{ {y}^{2} - \dfrac{ { \alpha }^{2}y }{ \beta } }}}}}$

Answer 2

$</p><p>\huge{\tt{\pink{Hello!!!}}}$

• An object is performing SHM ,

• Instantaneous Acceleration is β

• Instantaneous velocity is α

$</p><p>\tt{\red{ Relationship \:Between\: Amplitude \:, Acceleration\: And\: Velocity\: - }}$

$\mathfrak{\orange{Acceleration = - {\omega}^{2}y=\betaacc.=−ω}}$

$</p><p>\tt{\purple{Velocity=\omega\sqrt{{A}^{2} - {y}^{2}}=\alphavelocity = ω}}$

$</p><p>\tt{\blue{ Velocity\: in\: terms \:of \:angular\: frequency \: - }}$

$</p><p>\mathfrak {\orange{\beta = - \bigg \{\dfrac{ { \alpha }^{2} }{ {A}^{2} - {y}^{2} } \bigg \}}}$

$</p><p>\mathfrak {\red{= > {A}^{2} - {y}^{2} = - \dfrac{ { \alpha }^{2} y}{ \beta }}}$

$</p><p>\mathfrak {\green{= > {A}^{2} = {y}^{2} - \dfrac{ { \alpha }^{2}y }{ \beta }}}$

$</p><p>\tt{\orange{= > A = \sqrt{ {y}^{2} - \dfrac{ { \alpha }^{2}y }{ \beta } }}}</p><p>$

$</p><p>\fbox{ \purple{ \huge{ \bold{A = \sqrt{ {y}^{2} - \dfrac{ { \alpha }^{2}y }{ \beta } }}}}}$