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Grade: 12th Pass
        The magnitude of displacement of a particle moving in a circle of radius 'a' with constant angular speed 'w' varies with time 't' as:  
8 years ago

Answers : (2)

vikas askiitian expert
509 Points
							

R = acos@ (i) + asin@ (j)       ................1        

R is the position vector of particle at any time performing circular motion...

@ is angular displacement

 

w = @/t

@ = wt         ..............2

from 1 & 2

Rt = acoswt (i) + asinwt (j)          .............3

at t = 0 ,

Ro = a (i)             ..............4

 

displacement = Rt - Ro = a(coswt-1) (i) + a(sinwt) (j)

magnitude =[ (a2(coswt-1)2 + a2(sinwt)2]1/2

                =  [2a2 - 2a2coswt]1/2

                = a[2(1-coswt)]1/2

 

now , coswt = (1-2cos2(wt/2)) , after putting this in above eq

  magnitude = 2a[cos2(wt/2)]1/2

                  =2acos(wt/2)

this is the required result ,

 

approve if u like my ans

8 years ago
akvstr18
11 Points
							in the above answer it sayscoswt=1-cos^2(wt/2)but coswt = 1-sin^2(wt/2)=cos^2 (wt/2)-1also 2^(1/2) is not considered the final magnitude will be: (2)^(1/2)asin (wt/2)
						
one year ago
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