IIT JEE , AIEEE Forum - Physics , Mechanics

neeraj_agarwal_1990

A particle is moving in a circle of radius R=1m with constant speed v=4m/s.The ratio of displacement to acceleration of the foot of the perpendicular drawn from the particle on the diameter of the circle is ?

ans-1/16 (second)²

anchitsaini

y = a sin wt

d2y / dx^2 = a = - w^2 a sin wt

y/a = -1 / w^2

also v = rw which gives w = 4

hence

answer - 1/16

neeraj_agarwal_1990

how do we know that its an SHM question?

does this circular motion mean SHM??

anchitsaini

" foot of the perpendicular drawn from the particle on the diameter of the circle "

neeraj_agarwal_1990

i didn't understand this sentence...
does it mean that we have drawn perpendicular to a fixed diameter from the particle...?
how can we say that it executes shm...can't it be any oscillatory motion...

ganesha1991

anchit is right
the projection of the particle on the diameter will execute shm

neeraj_agarwal_1990

won't we have to prove that its shm??

how can we say it just by observing?

anchitsaini

from the figure it can be clearly seen that --

projection of foot of perpendicular on vertical diameter =

y = A sin @

also

@ = wt +

t² / 2

but

= 0(no angular acceleration)

hence

@ = wt

thus

y = A sin wt

which equation of s.h.m

neeraj_agarwal_1990

thanx....but is it some standard shm??

u just used it without proving it in the first post...

anchitsaini

ya
it is standard
indeed the very starting part of chapter shm involves the relation between
uniform circular motion and shm

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