# Don't understand with the question

1. Oct 30, 2004

### Sanosuke Sagara

This is a simple harmonic motion question.Two particles are in simple harmonic motion along a straight line of length 20.0cm.The period of each particle is 1.5s but there is a phase difference of 30 degree between them.

Just what it means by phase difference of 30 degree ?

Can anybody explain this to me ? Thank you .

2. Oct 30, 2004

### allistair

it means that the second particle reaches it's "amplitude" (not sure how to translate amplitude to english but it should be something similar) 30/360 * 1.5s later (or earlier than the first)

3. Oct 31, 2004

### Sanosuke Sagara

to confirm whether my solution is right or wrong

After this, a question ask like this,

What is the distance between the two particles 0.5s after the first particle passes the equilibrium point ?

My solution is :

30/360 * 1.5s = 0.125s

1.5 + 0.125 =1.625s (Period of the second particle)

Angular velocity for the second particle,

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4. Nov 1, 2004

### Sanosuke Sagara

Please, someon please help me.My solution is in the attachment.Just need someone to confirm whether my soulution to the question is right or wrong. Thanks for the help !!!!

5. Nov 1, 2004

### Galileo

Hi Sanosuke.

If you set up the equation of motion for both particles:
$$x_1(t)=20\sin(\omega t)$$
$$x_2(t)=20\sin(\omega t-30)$$
(With the arguments are in degrees, not radians)
You are given the period is 1.5 s, so $\omega=\frac{360}{1.5}=240 deg/s$.
Then $x_1[/tex] passes the equilibrium point at [itex]t=0$
You are asked for the distance [itex]d=|x_1-x_2|when t=0.5 s.

I get d=2.68 cm.

6. Nov 1, 2004

### Sanosuke Sagara

Anyway thanks for your help in this question . I appreciate it.