JEE MAIN - Physics (2021 - 24th February Evening Shift - No. 9)
When a particle executes SHM, the nature of graphical representation of velocity as a function of displacement is :
circular
straight line
parabolic
elliptical
Explanation
Since, the particle is executing SHM.
Therefore, displacement equation of wave will be
$$y = A\sin \omega t$$
$$ \Rightarrow y/A = \sin \omega t$$
and wave velocity equation will be
$${v_y} = {{dy} \over {dt}} = A\omega \cos \omega t$$
$$ \Rightarrow {v_y}/A\omega = \cos \omega t$$
Now, $${\sin ^2}\omega t + {\cos ^2}\omega t = 1$$
$$\therefore$$ $${(y/A)^2} + {({v_y} / A\omega )^2} = 1$$
This equation is similar to the equation of ellipse.
Therefore, displacement equation of wave will be
$$y = A\sin \omega t$$
$$ \Rightarrow y/A = \sin \omega t$$
and wave velocity equation will be
$${v_y} = {{dy} \over {dt}} = A\omega \cos \omega t$$
$$ \Rightarrow {v_y}/A\omega = \cos \omega t$$
Now, $${\sin ^2}\omega t + {\cos ^2}\omega t = 1$$
$$\therefore$$ $${(y/A)^2} + {({v_y} / A\omega )^2} = 1$$
This equation is similar to the equation of ellipse.
Comments (0)
