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Grade: 11

                        

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) 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..

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..

9 years ago

Answers : (6)

vikas askiitian expert
509 Points
							

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 ....

9 years ago
Sudheesh Singanamalla
114 Points
							

This Question has more than 1 answer

answer is (A,B,C)

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

 

Please approve !!

9 years ago
Mayank Kumar Jha
19 Points
							

my dear sudhesh the focus of the ellipse is not the center and that is what i wish to highlgiht..

9 years ago
Sudheesh Singanamalla
114 Points
							

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.

 

Please approve !

9 years ago
Mayank Kumar Jha
19 Points
							

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...

9 years ago
Sahil shaikh
18 Points
							
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
6 years ago
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