IIT JEE , AIEEE Forum - Physics , Mechanics

abhijeet_0201

this one's frm bmat fst 3 paper 1

i can t understand the sol either so plz help

two particles 1 and 2 are in shm with same amplitude A and same angular vel w

at time t=0 one is at x=A/2 and the other is at (-A/

2).both are moving in the same direction they will collide after time t=

(A) 5T/12 (B)5T/24 (C)5T/16 (D)5T/8

adjoining diagram

<-------- <---------

-----------------x-------------.------------x------------

2 c 1

A/root2 A/2

abhijeet_0201

hello?????????????

karthik2007

I am not quite getting the answer.

The two motions can be described by

x1 = Asin(wt + pi/6)

and

x2 = Asin(wt-pi/4)

On solving, I am getting something else :(

bhuwanaroracorroded

gettin13/48T

abhijeet_0201

i have added the adjoining diagram(was given along with ques) if it helps u understand the ques better

shall i post the sol maybe u can help me understand it .....

ps:any bt student can look up the sol and try to exp me!!!!!!!

karthik2007

Ya post the solution. I'll try to explain it through the sol

abhijeet_0201

OKAY SO HERE'S THE SOL
I CANT MAKE THE DIAG(THEY HAVE CONSIRED SHM AS REFLEC ON CIRCLE)
SO ILL JUST TYPE THE SOL

the phase diff b/w the particles
phi=pi/6 +pi/4 =5pi/12

th will collide when the perpendicular s drawn on the the diameter coincide.this will happen when any of the particle rotates an angle

theta =phi +phi' (on the figure they ,ve shown phi''=5pi/24 (dont know how)

=5pi/12 + 5pi/24=5pi/8
or wt =5pi/8
t= 5pi/8 x T/2pi

=5T/16'

karthik2007

draw a diagram in paint

rooney

Answer should be 13T/48.

In calculating theta, they have taken phi = 5 pi/12 where as it should be pi/3 or 60 degrees. Then, theta = 13pi/24. Dividing by w=2pi/T, we get t = 13T/48.

elessar_iitkgp

The answer is incorrect.

Two particles are moving in a circle with constant angular velocity. The first conclusion that should have struck you is that their angular separation will remain constant. The projection of their circular motions on a straight line is SHM.

Now, given the conditions about the initial positions, you should be able to calculate the phase angles. They are the angles that the particles are making with the horizontal line on which the circular motion is being projected. See the diagram below.

As the particles rotate, the angular separation between them is constant, and that is 5

/12.

Now, they on the line at which the circular motion is being projected, the positions of the particles is same only if the perpendiculars from the particles to the diameter lie in a straight line. See the second diagram. At such a position, the line joining the two particles is perpendicular to the diameter. Now, from geometry, you can say that the diameter bisects the angle formed by the two particles. So the angles made by the particles with the diameter is 5

/24.

So in this time from the starting the first particle has traveled an angle of

-

/4 - 5

/24
= 13

/24

So

t₀ = 13

/24
t₀ = 13

/24

= (13

/24)/(2

/T) = 13T/48

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