#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# 26. 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 (B) the velocity and acceleration of the particle are normal to each other at t = p/2p (C) the acceleration of the particle is always directed towards a focus (D) the distance travelled by the particle in time interval t = 0 to t = p/2p is a.  my doubt is that option c should be wrong... in the answer key its given to be true which says that the focus is the centre which in no way is visible to me plz confirm..

10 years ago

x = acospt                         y = bsinpt

R(position vector) = acospt(i) + bsinpt(j)        ..................1

dx/dt = vx = -apsinpt           dy/dt = vy =  bpcospt

v = -apsinpt(i) + bpcospt (j)

dv/dt = ax = -ap2cospt         dv/dt = ay = -bp2sinpt

a = -ap2cospt(i) - bp2sinpt(j)

a = -p2( acospt (i) + bsinpt (j) )      .....................2

from eq 1

a = -p2(R)

R is always directrd towards center (0,0) so accleration

of particle will always be towards center of ellipse ....

10 years ago

This Question has more than 1 answer

x =  a cos pt => cos pt = x/a

y = b sin pt => sin pt = y/b

squaring and adding them we get

x2/a2 + y2/b2 = 1

Therefore path is an ellipse --------------- (option A is correct)

dx/dt = vx = -a p sin pt

d2x/ dt2 = ax = - ap2 cos pt

dy/dt = vy = bp cos pt

d2y/dt2 = - bp2 sin pt

At time t = pi/2p or pt = pi/2

ax and vy become 0

only vx and vy are left

or we can say velocity is along the negative x axis and accelaration along the negative y axis

at t = pi/2p , velocity and accelaration are normal to each other , so option B is also correct.

At t=t , position of the particle

r-> = x i^ + y j^ = a cos pt i^ + b sin pt  j^

and accelaration 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)

so the accelaration of the particle is always directed towards origin

hence option (C) is also correct

10 years ago

Dear Mayank ,

if you read carefully i said it is towards the origin not the focus , so , i m really sorry , option (C) is wrong , i thought the question was about particle always directed towards the origin , i didn't see the 'focus' part.

so option (A) is correct and option (B) is correct.

10 years ago

yaar sudhesh option c says accn. is directed towards a focus wheras uve proved it to be directed towards te fosuc and seeing the trajectory of particle the focus in no way is the center of the ellipse...

6 years ago
x =  a cos pt => cos pt = x/a
y = b sin pt => sin pt = y/b
squaring and adding them we get
x2/a2 + y2/b2 = 1
Therefore path is an ellipse --------------- (option A is correct)
dx/dt = vx = -a p sin pt
d2x/ dt2 = ax = - ap2 cos pt
dy/dt = vy = bp cos pt
d2y/dt2 = - bp2 sin pt
At time t = pi/2p or pt = pi/2
ax and vy become 0
only vx and vy are left
or we can say velocity is along the negative x axis and accelaration along the negative y axis
at t = pi/2p , velocity and accelaration are normal to each other , so option B is also correct.
At t=t , position of the particle
r-> = x i^ + y j^ = a cos pt i^ + b sin pt  j^
and accelaration 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)
so the accelaration of the particle is always directed towards origin
hence option (C) is also correct