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

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Solution

The correct option is **D** the acceleration of the particle is always directed towards a focus

x=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.

x=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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