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  • A particle is acted simultaneously by mutually perpendicular simple harmonic motion x = a cos ωt and y = a sin ωt. The trajectory of motion of the particle will be

A particle is acted simultaneously by mutually perpendicular simple harmonic motion x = a \cos \omega t and y = a \sin \omega t. The trajectory of motion of the particle will be
(a) an ellipse

(b) a parabola

(c) a circle

(d) a straight line

Answers (1)

The answer is the option (c) a circle.

Explanation: We know that,

Resultant displacement = x + y

                                        = a \cos \omega t + a \sin \omega t

Thus, y^{'} = a \left (\cos \omega t + a \sin \omega t \right )

                  =a \sqrt{2}\left [ \frac{\cos \omega t}{\sqrt{2}}+ \frac {\sin \omega t}{\sqrt{2}} \right ]

                      =a \sqrt{2} \left [ \cos \omega t \cos 45^{\circ} +\sin \omega t \sin 45^{\circ} \right ]

……. (particle is acted simultaneously by mutually perpendicular direction)

y^{'}= a\sqrt{2}\cos s\left ( \omega t - 45^{\circ} \right )     ……. Thus, the displacement can neither be a straight line nor a parabola

Now, let us square and add x & y,

x^{2} + y^{2} = a^{2} \cos^{2} \omega t + a^{2} \sin^{2} \omega t

Thus, x^{2} + y^{2} = a^{2}

This is the equation of a circle, hence, opt (c)

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