The coordinates of a particle moving in a plane are given by x(t) = a cos(πt) and y(t) = b sin(πt) , a,b (

Question

The coordinates of a particle moving in a plane are given by \( x(t)=a \cos (\pi t) \) and \( y(t)=b \sin (\pi t) \), where \( a, b(<a) \) and \( p \) are positive constants of appropriate dimensions then - [JEE 1999] (1) the path of the particle is an ellipse (2) the velocity and acceleration of the particle are normal to each other at \( t=\pi / 2 \pi \) (3) the acceleration of the particle is always directed towards a focus \( \rightarrow R B D \) (4) the distance travelled by the particle in time interval \( t=0 \) to \( =\pi / 2 \pi \) is a.

Answer 1

Correct option : (1), (2), (3)

Answer 2

Correct options are (1), (2) and (3)

\(x = a\cos (pt) \Rightarrow \cos (pt) = \frac xa\quad....(i)\)

\(y = b\sin (pt) \Rightarrow \sin(pt) = \frac yb\quad....(ii)\)

Squaring and adding Eqs. (i) and (ii), we get

\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)

Therefore, path of the particle is an ellipse. Hence, option (a) is correct.

From the given equations we can find

\(\frac {dx}{dt} = v_x = -ap \sin pt\)

\(\Rightarrow \frac{d^2 x}{dt^2} = a_x = -ap^2\cos pt\)

\(\frac {dy}{dt} = v_y = bp \cos pt\)

\(\Rightarrow \frac{d^2 y}{dt^2} = a_y = -bp^2\sin pt\)

At time t = π/2p or pt = π/2

a_x​ and \(v - \nu\) become zero (because cos π/2 = 0)

only v_x​ and a_y​ are left.

or we can say that velocity is along negative x-axis and acceleration along y-axis.

Hence, at t = π/2p velocity and acceleration of the particle are normal to each other. So, option (b) is also correct.

At t = t, position of the particle

\(r(t) = x\hat i + y\hat j = a\cos pt\hat i + b \sin pt\hat j\)

and acceleration of the particle is

\(a(t) = a_x\hat i + a_y\hat j = - p^2 [a \cos pt\hat i + b\sin pt \hat j]\)

\(= -p^2[x\hat i + y\hat j]\) 

\(= -p^2 r (t)\)

Therefore, acceleration of the particle is always directed towards origin.

Hence, option (c) is also correct.

At t = 0, particle is at (a, 0) and at t = π/2p particle is at (0, b).

Therefore, the distance covered is one-fourth of the elliptical path not a.

Hence, option (d) is wrong.