First slide

Simple harmonic 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 $x = A \sin (ω t)$ and $y = A \sin (ω t + \frac{π}{6})$ respectively. The trajectory followed by the charged particle is

Moderate

A

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

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

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

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

Solution

When two SHM's (which are perpendicular to each other) are given by $x = A_{1} ω t$ and $y = A_{2} \sin (ω t + δ)$ and when they are combined the trajectory of the particle is given by
$\frac{x^{2}}{A_{1}^{2}} + \frac{y^{2}}{A_{2}^{2}} - \frac{2 x y}{A_{1} A_{2}} \cos δ = \sin^{2} δ$
(This is equation of ellipse)
In the given equations, phase difference is $δ = \frac{π}{6}$ and $A_{1} = A_{2} = A$
$\begin{array}{l} ∴ \frac{x^{2}}{A_{}^{2}} + \frac{y^{2}}{A_{}^{2}} - \frac{2 x y}{A^{2}} \cdot \frac{\sqrt{3}}{2} = \frac{1}{4} \\ x^{2} + y^{2} - \sqrt{3} x y = \frac{1}{4} A^{2} and this is an ellipse. \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App