MY CART (5)

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

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R

There are no items in this cart.
Continue Shopping
Menu
Get instant 20% OFF on Online Material.
coupon code: MOB20 | View Course list

  • Complete Physics Course - Class 11
  • OFFERED PRICE: R 2,800
  • View Details
Get extra R 2,800 off
USE CODE: chait6

				   

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..

5 years ago

Share

Answers : (6)

										

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 ....

5 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


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 !!

5 years ago
										

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

5 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 !

5 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...

5 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
2 years ago

Post Your Answer

Other Related Questions on Mechanics

a packet is released from a balloon accelerating upward with acceleration a. the acceleration of packet just after the release is
 
 
Since the ballon was moving upward with acceleration a, there is a force of gravity on the packet too. The balloon is moving upward bexause the reaction force R>W, weight of the packet. As...
 
Shaswata Biswas 3 months ago
 
Just after the release the packet posses only the component of the velocity aquired. Here the accleration of the balloon has no effect on the acceration of the body. When the body is...
 
Shaswata Biswas 3 months ago
 
The accleration of the packet will be `g` acting downward as no external force other then gravity working on it.it will achieve the case of free fall. If I am wrong please ..explain the...
 
Chandan kumar mandal 2 months ago
The potential energy of a particle of mass 1kg moving along x-axis is given by U(x)=[x^2/2-x]J. If total mechanical energy of the particle is 2J find its maximum speed.
 
 
total mechanical energy = U+ K.E to attain maximum speed the obect must have maximum K.E as K.E will be maximum , U has to be minimum (conservation of energy )given - U(x) = x²/2-xfor this...
 
fizaparveen 2 months ago
 
We know that the object accelerate till force is applied on it and it attains maximum velocity just after the force becomes zero... SO we know the negative of potential energy gradient is...
 
Suraj Singh 20 days ago
 
Ans- 2Speed will be max when kinetic energy is maximum so potential energy will be minimum so differentiating the function put it equal to 0 so we obtain minimum for x=0 now kin potential...
 
rishabh doshi 2 months ago
if a body move with u velocity then after collide with wall and remove with v velocity what is the impulse
 
 
Hi aman, I think this might help as we know that Impulse and also F=m(dv/dt) so (here m is the mass of body and dv/dt =acc) here p is the momentum ans delta p is change in momentum hence in...
 
Ankit Jaiswal 3 months ago
what is work energy thereom...................................?
 
 
The work-energy theorem is a generalized description of motion which states that work done by the sum of all forces acting on an object is equal to change in that object’s kinetic energy....
 
dolly bhatia one month ago
 
The work W done by the net force on a particle equals the change in the particle`s kinetic energy KE: W = Δ K E = 1 2 m v f 2 − 1 2 m v i 2 . The work-energy theorem can be derived from...
 
Sushmit Trivedi one month ago
 
For any net force acting on a particle moving along any curvilinear path, it can be demonstrated that its work equals the change in the kinetic energy of the particle by a simple derivation...
 
rahul kushwaha 3 months ago
The cieling of long hall is 25 m high What is the maximum horizontal distance that a ball thrown with a speed of 40 ms can go without hitting the ceiling of wall? Plz explain with a DIAGRAM
 
 
Let the angle of throwing be Q with respect to the ground So, the range is R = 40*cosQ*t where t = time taken reach ground. Now, we know the time taken to reach the top of its flight is half...
 
Tapas Khanda 5 months ago
 
Let the angle of throwing be Q with respect to the ground So, the range is R = 30*cosQ*t where t = time taken reach ground. Now, we know the time taken to reach the top of its flight is half...
 
Tapas Khanda 5 months ago
 
Diagram is not required to solve this question. You can directly use the formula to find range : R = (v^2/g)*sin2Q Maximum height reached, H = v^2*sin^2Q/2g So, 25 = 40^2*sin^2Q/(2*9.8) So,...
 
Tapas Khanda 5 months ago
What harmfull gases are produced when we burn plastics? And can we burn plastics?
 
 
I`m doing a project, can someone help me, How to reduce those toxic contents when plantic are burnt Like sulphur, nitrogen etc
 
2017 years ago
View all Questions »

  • Complete Physics Course - Class 12
  • OFFERED PRICE: R 2,600
  • View Details
Get extra R 2,600 off
USE CODE: chait6

  • Complete Physics Course - Class 11
  • OFFERED PRICE: R 2,800
  • View Details

Get extra R 2,800 off
USE CODE: chait6

More Questions On Mechanics

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!!
Click Here for details