Question

JEE questions for me​

Answer 1

$\Huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

Displacement in SHM

$\large{\tt{x = A \: sin( \omega \: t)}} \\ \\ \large{\longrightarrow \: \tt{x = A \: \sqrt{1 - {cos}^{2}( \omega \: t)} - - - - (1) }}$

Velocity of a particle in SHM is given by

$\large{\tt{v = A \omega \: cos( \omega \: t) - - - - (2)}}$

From the above given equations,we deduce the relation:

$\boxed{ \boxed{ \large{\tt{ {v}^{2} = \omega \sqrt{ {A}^{2} - {x}^{2} }}}}}$

Now,

$\hookrightarrow \ \sf{ \dfrac{v}{ \omega} = \sqrt{ {A - {x}^{2} }^{2} } } \\ \\ \hookrightarrow \: \sf{ \frac{ {v}^{2} }{ \omega} + x {}^{2} = {A}^{2} } \\ \\ \huge{ \hookrightarrow \: \sf{ \frac{ {v}^{2} }{(A \omega) {}^{2} } + \frac{ {x}^{2} }{ {A}^{2}} = 1 }}$

The above equation is of the form :

$\sf{\frac{ {x}^{2} }{ {a}^{2}} \: + \frac{y {}^{2} }{b {}^{2}} = 1}$

• This is an equation for an ellipse

Plotting v - x graph would give an elliptical shape on the graph

NOTE

Here,

• A is the Amplitude of particle

• W is the Angular Velocity

• V is the Velocity

• X is the Displacement

Answer 2

Answer:

since the question is of JEE examination so it will be time consuming to jump for solutions ( which is not required ) instead of getting to the answer ( which is required )

the correct answer is "the graph is of ellipse"