Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

A particle moves in the x-y plane according to rule \[x = a\cos \omega t\] and \[y = a\sin \omega t\]. The particle follows
(A) An elliptical path
(B) A circular path
(C) A parabolic path
(D) A straight line path inclined equally to x and y axes.

seo-qna
Last updated date: 17th Jun 2024
Total views: 52.8k
Views today: 0.52k
Answer
VerifiedVerified
52.8k+ views
Hint: The equations given are parametric equations of a particular shape. We need to compare the equation with the parametric equations of the given options.
Formula used: In this solution we will be using the following formulae;
\[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\] Cartesian equation of an ellipse, with semi major axis \[a\] and semi minor axis \[b\]. \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{a^2}}} = 1\] Cartesian equation of a circle, with radius \[a\].
\[x = a\cos \omega t\] and \[y = b\sin \theta \] this is the parametric equation of an ellipse. \[x = a\cos \theta \] and \[y = a\sin \theta \] this is the parametric equation of a circle

Complete Step-by-Step Solution:
A particle is said to move in the x y plane according to the rule
\[x = a\cos \omega t\] and also \[y = a\sin \omega t\]
This is a parametric equation. We shall compare it to the parametric equation of each of the options.
Option A is given an ellipse. The parametric equation of an ellipse centred at the origin is given as
\[x = a\cos \omega t\] and \[y = b\sin \theta \] where \[a\] is the semi major axis and \[b\] is the semi minor axis. (this can be interchanged, depending on the orientation of the ellipse)
Hence, by observation, we see that this is not the equation is not that of an ellipse.
Option B is given as a circle. The parametric equation of a circle is given as
\[x = a\cos \theta \] and \[y = a\sin \theta \] where \[a\] is the radius.
Hence by observation, we can see that the equation above is exactly like the equation of a circle where \[\theta = \omega t\]
Hence, the particle traces out a circular path

Thus, the correct option is B.

Note: Alternatively, the equation can be converted to Cartesian form or standard form. This can be done as follows,
 Squaring both x and y equation, we have
\[{x^2} = {\left( {a\cos \omega t} \right)^2} = {a^2}{\cos ^2}\omega t\], and for y coordinate,
\[{y^2} = {\left( {a\sin \omega t} \right)^2} = {a^2}{\sin ^2}\omega t\]
Now, we add the two equations together, such that we get
\[{x^2} + {y^2} = {a^2}{\cos ^2}\omega t + {a^2}{\sin ^2}\omega t\]
By factorising out \[{a^2}\], we get
\[{x^2} + {y^2} = {a^2}\left( {{{\cos }^2}\omega t + {{\sin }^2}\omega t} \right)\]
Then, since\[{\cos ^2}\omega t + {\sin ^2}\omega t = 1\], we have
\[{x^2} + {y^2} = {a^2}\] or
\[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{a^2}}} = 1\]
And this is precisely the Cartesian equation of a circle.
An ellipse would have been,
\[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\]