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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..

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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