Phase difference of two moving particle

[desmond iking]

Homework Statement

the question doesn't state that which particle is leading/lagging which particle.

Homework Equations

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?

 

[tms]

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

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.

 

[desmond iking]

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

 

[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).

 

[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?

 

[tms]

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

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

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

Correct.

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

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.

how to solve it without the help of the graphs?

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.