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` `**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..

6 years ago

x = acospt y = bsinpt

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

dx/dt = v

_{x}= -apsinpt dy/dt = v_{y}= bpcosptv = -apsinpt(i) + bpcospt (j)

dv/dt = a

_{x}= -ap^{2}cospt dv/dt = a_{y}= -bp^{2}sinpta = -ap

^{2}cospt(i) - bp^{2}sinpt(j)a = -p

^{2}( acospt (i) + bsinpt (j) ) .....................2from eq 1

a = -p

^{2}(R)R is always directrd towards center (0,0) so accleration

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

6 years ago

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

x

^{2}/a^{2}+ y^{2}/b^{2}= 1Therefore path is an ellipse --------------- (option A is correct)

dx/dt = v

_{x}= -a p sin ptd

^{2}x/ dt^{2}= a_{x}= - ap^{2}cos ptdy/dt = v

_{y }= bp cos ptd

^{2}y/dt^{2}= - bp^{2}sin ptAt time t = pi/2p or pt = pi/2

a

_{x}and v_{y}become 0only v

_{x}and v_{y}are leftor 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) = a_{x}i^{^ }+ a_{y}j^{^}= - p

^{2}[ a cos pt i^{^}+ b sin pt j^{^}]= - p

^{2}[ x i^{^}+ y j^{^}]= - p

^{2 }r^{->}(t)so the accelaration of the particle is always directed towards origin

hence option (C) is also correct

Please approve !!

6 years ago

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

6 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.

Please approve !

6 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/ay = b sin pt => sin pt = y/bsquaring and adding them we getx2/a2 + y2/b2 = 1Therefore path is an ellipse --------------- (option A is correct)dx/dt = vx = -a p sin ptd2x/ dt2 = ax = - ap2 cos ptdy/dt = vy = bp cos ptd2y/dt2 = - bp2 sin ptAt time t = pi/2p or pt = pi/2ax and vy become 0only vx and vy are leftor we can say velocity is along the negative x axis and accelaration along the negative y axisat t = pi/2p , velocity and accelaration are normal to each other , so option B is also correct.At t=t , position of the particler-> = x i^ + y j^ = a cos pt i^ + b sin pt j^and accelaration of the particle isa-> (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 originhence option (C) is also correct

2 years ago

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