Question

# The coordinates of a particle moving in a plane are given by x(t) = a cos (pt) and y(t) = b sin (pt) where a, b (<a) and p are positive constants of appropriate dimensions. Then

A
the path of the particle is an ellipse
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B
the velocity and acceleration of the particle are normal to each other at t=π/(2p)
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C
the acceleration of the particle is always directed towards a focus
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D
the distance travelled by the particle in time interval t=0tot=π/(2p) is a
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Solution

## The correct option is D the acceleration of the particle is always directed towards a focusx=a cos pt ⇒ cos(pt)=xa ...(i) y=b sin pt ⇒ sin(pt)=yb ...(ii) Squaring and adding (i) and (ii), we get x2a2+y2b2=1 ∴ Path of the particle is in ellipse. Hence option (a) is correct. From the given equations We can find, dxdt=vx=−ap sin pt d2x=ax=−ap2 cos pt dydt=vy=bp cos pt and d2ydt=ay=−bp2 sin pt At time t=π2p or pt = π2 ax and vy become zero (because cos π2=0) only vx and ay are left. or we can say that velocity is along negative x-axis and acceleration along y-axis. Hence, at t=π2p, velocity and acceleration force the particle are normal to each other. So option (b) is also correct. At t = t, position of the particle →r(t)=x^i+y^j= a cos pt^i+b sin pt^j and acceleration of the particle is →a(t)=ax^i+ay^j=p2[a cos pt^i+b sin pt^j] =p2[x^i+y^j=−p2→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=π2p, particle is at (0, b). Therefore, the distance covered is one fourth of the elliptical path not a. Hence option (d) is wrong.

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