A particle is acted simultaneously by mutually perpendicular simp… - Vidyadip

MCQ

A particle is acted simultaneously by mutually perpendicular simple hormonic motions $\text{x}=\text{a}\cos\omega\text{t}$ and $\text{y}=\text{a}\sin\omega\text{t}.$ The trajectory of motion of the particle will be:

A
An ellipse.
B
A parabola.
✓
A circle.
D
A straight line.

✓

Answer

Correct option: C.

A circle.

Resultant displacement is $x + y$
$\text{x}=\text{a}\cos\omega\text{t}\ ...(1)$
$\text{y}=\text{A}\sin\omega\text{t}\ ...(2)$
Dispacement $=\text{a}\cos\omega\text{t}+\text{a}\sin\omega\text{t}$
$\text{y}'=\text{a}\sqrt{2}\Big[\frac{\cos\omega\text{t}}{\sqrt{2}}+\frac{\sin\omega\text{t}}{\sqrt{2}}\Big]$
$\text{y}'\text{a}\sqrt{2}\ [\cos\omega\text{t}\cos45^\circ+\sin\omega\text{t}\sin45^\circ]$as the particle is acted simultaneously by Mutually perpendicular direction.
$\text{y}'=\text{a}\sqrt{2}\cos\text{s}(\omega\text{t}-45^\circ)$
Hence the displacement is neither a straight line nor a parabola.
Now, squaring and adding $(i), (ii)$
$\text{x}^2+\text{y}^2=\text{a}^2\cos^2\omega\text{t}+\text{a}^2\sin^2\omega\text{t}=\text{a}^2[\cos^2\omega\text{t}+\sin^2\omega\text{t}]$
$\text{x}^2+\text{y}^2=\text{a}^2$
This shows the equation of circle, hence the motion is circular motion.

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