Homework Help: Phase difference of two moving particle

1. Jun 21, 2014

desmond iking

1. The problem statement, all variables and given/known data

the question doesnt state that which particle is leading/lagging which particle.

2. Relevant equations

3. The attempt at a solution

for the first particle, my working is 10 sin(2pi/3 )= 8.66cm

for the second paticle, my working is 10 sin(2pi/3+ 30pi/180)= 5 cm
so my final answer is 8.66-5 = 3.66

the sample ans is attached below.

how do we know that the second particle is lagging behind first particle?

Attached Files:

File size:
19 KB
Views:
118
• IMG_20140622_085946[1].jpg
File size:
26.8 KB
Views:
116
2. Jun 21, 2014

tms

I think it's just a question of language: the first particle is the leading particle simply because the leader is first by definition. The question could have been written more clearly.

I assume you have noticed that the given answer to the second part is obviously wrong.

Last edited: Jun 21, 2014
3. Jun 27, 2014

desmond iking

how do we know that both particles are moving away form moving closer to each other?

4. Jun 27, 2014

tms

In general, you look at the velocity of each. In this case, you can tell the velocity of one of the particles by inspection (using the numerical results from the answer sheet).

5. Jun 30, 2014

desmond iking

v_1 = 10 w cos(wt) = 10 (5/3 pi) cos(4/3 pi x 0.5) = -26.2 cm/s

v_2 = 10 w cos(wt-pi/6) = 10(5/3 pi) cos(4/3 pi x 0.5 - pi/6) = 0 cm/s.

my foundation for positive and negative velocity, and acceleration isn't good.
v1 is negative what does it mean?

how to solve it without the help of the graphs?

Last edited: Jun 30, 2014
6. Jun 30, 2014

tms

That should be $4\pi/3$, not $5\pi/3$, but generally correct.

Correct.

It means that the direction is opposite to the direction of positive velocity. Assuming $x$ increases to the right, then $v$ also increases to the right, and a negative velocity is to the left.

I'm not sure what you mean, since you just did solve it without a graph. If you are asking what I meant when I said one of the velocities could be found by inspection, the second particle's position was at $x = 10$, and you know from the problem statement that the amplitude of the motion was also $10$. So that particle was at the extreme of its motion, so its velocity had to be zero at that moment.