First slide

Simple hormonic motion

Question

A charged particle is deflected by two mutually perpendicular oscillating electric fields such that the displacement of the particle due to each one of them is given by $x = A \sin (ωt) and y = A \sin (ωt + \frac{π}{6})$ respectively. The trajectory followed by the charged particle is

Difficult

A

a circle with equation $x^{2} + y^{2} = A^{2}$
B

a straight line with equation $y = \sqrt{3} x$
C

an ellipse with equation $x^{2} + y^{2} - xy = \frac{3}{4} A^{2}$
D

an ellipse with equation $x^{2} + y^{2} - \sqrt{3} xy = \frac{1}{4} A^{2}$

Solution

When two SHM's (which are perpendicular to each other ) are given by $x = A_{1} \sin (ω t)$ and $y = A_{2} \sin (ωt + δ)$ and when they are combined, the trajectory of the particle is given by
$\frac{x^{2}}{{A_{1}}^{2}} + \frac{y^{2}}{{A^{2}}_{2}} - \frac{2 xy}{A_{1} A_{2}} cosδ = \sin^{2} δ$ (This is equation of ellipse).
For the given problem $A_{1} = A_{2} = A$ and $δ = \frac{π}{6}$ ,
$\frac{x^{2}}{A^{2}} + \frac{y^{2}}{A^{2}} - \frac{2 xy}{A^{2}} . \frac{\sqrt{3}}{2} = \frac{1}{4}$
$x^{2} + y^{2} - \sqrt{3} xy = \frac{1}{4} A^{2} and this is an ellipse$

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