Question

Two particles are in SHM in a straight line. Amplitude 'A' and time period 'T' of both the particles are equal. At time t=0 , one particle is at displacement y1=+A and the other at y2=-A/2 and they are approaching towards each other. After what time they cross each other?

Ans. T/6

Answer 1

am not too good with this,can we try energy conservation ?

Answer 2

i tried something like \(y_1=A\sin(\frac{2 \pi t}{T}+\theta_1),y_1=A \\ y_2=A\sin(\frac{2 \pi t}{T}+\theta_2),y_2=-A/2\)
and when they cross each other their phases will be same, so, \(\theta_1=\theta_2\)
solving those three equations, i am not getting t=T/6....maybe you try again...

Answer 3

oh, got it.
\(\large y_1=A\sin(\frac{2 \pi [t-t_0]}{T}+\theta_1),y_1=A \\ \large y_2=A\sin(\frac{2 \pi [t_0-t]}{T}+\theta_2),y_2=-A/2\)
where t0=0.
now if u solve those, u get t=T/6

Answer 4

with \(\theta_1=\theta_2\) ofcourse.

Answer 5

@shubhamsrg
The point in energy conservation is that time disappears from the equations.
In this problem, time is the unknown, so you cannot use that method.

Answer 6

cant we express velocity in terms of time ?

Answer 7

Of course we can, but that would be a waste of time here, since the condition required is equal displacement.

Answer 8

The second body is late by T/3.

Let's write :
\(\large y_1=A\cos(\Large\frac{2 \pi t}{T}) \\ \large y_2=A\cos(\Large\frac{2 \pi [t-T/3]}{T})\)

You can check that at t = 0, y1 = A , and y2 = –A/2 and is increasing.

Least value of t for y1 = y2 leads to T/6

Answer 9

One can also solve it by using the method of projecting uniform circulur motion on one of the coordinate axis.
This method relieves us of the trigonometric calculations and makes it more of a geometrical problem.
Some complex problems are made very easy by this method.

Answer 10

hmm,,so much knowledge in this world and so less i know! :/