# Homework Help: Graph of a particle's position

1. Apr 19, 2015

### goonking

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

2. Relevant equations

3. The attempt at a solution
the particle is oscillating

so we know at 1.5 seconds, it is at the equilibrium point , where speed is highest. (the speeds are 0 at the highest and lowest points of the graph.)

we see that it goes from -6m to 6m, which is 12 meters in 1 second.

so is the answer 12 meters per second? since that is the highest it can go?

2. Apr 19, 2015

### AlephNumbers

Are you familiar with the equations that govern simple harmonic motion?

3. Apr 19, 2015

### goonking

no sir, i will google it right now.

4. Apr 19, 2015

### SammyS

Staff Emeritus
That's just the average speed from t = 1 second to t = 2 seconds .

5. Apr 19, 2015

### AlephNumbers

6. Apr 19, 2015

### goonking

7. Apr 20, 2015

### goonking

8. Apr 20, 2015

### AlephNumbers

Yes.

9. Apr 20, 2015

### goonking

how do we find angular velocity? Is the particle moving around in a circle? I thought it was more like a spring, going back and forth.

10. Apr 20, 2015

### AlephNumbers

It is.

No, but its horizontal displacement is the same as if it were moving in uniform circular motion. I would look at a textbook or online resource for a more complete explanation.

The distance between two crests (high points) on the graph you provided is the amount of time it takes the mass to make a complete "revolution" (2π radians). Angular velocity is measured in radians per second... do you see where I am going with this?

11. Apr 20, 2015

### goonking

yes, so it takes 2 seconds to do 1 rev.

ω = 1 radian per sec.

A = 3 meters

and cos ω t = cos ((1r/s) (2 second)) = .99939

v = ω A cos ωt = 1 x 3 x .999 = 3 m/s

correct?

12. Apr 20, 2015

### AlephNumbers

On a bit of a physics binge, eh? Don't worry, PF has you covered.

13. Apr 20, 2015

### BvU

Draw a straight line with slope 3 m/s through the point (1.5s, 0 m) to check...

14. Apr 20, 2015

### goonking

yes, silly me always doing homework when everyone is asleep.

15. Apr 20, 2015

### BvU

Well done !

16. Apr 20, 2015

### goonking

whoops, yes.

and cos ω t = cos ((π r/s) (2 second)) = .9984

answer still came out to be 3 m/s.

17. Apr 20, 2015

### goonking

like this?

I have no idea how to confirm if that is suppose to correct or not.

18. Apr 20, 2015

### AlephNumbers

No, I think BvU meant more like this

19. Apr 20, 2015

### goonking

i think he was trying to tell me my answer was wrong.

by the looks of it, my speed should be -3m/s instead of 3 m/s.

correct?

20. Apr 20, 2015

### AlephNumbers

You just drew a tangent line at t = 2.5 instead of t = 1.5. At t = 2.5 the mass has a negative velocity. I believe BvU was just pointing out that you could draw a line tangent to the graph in order to check your answer.

Last edited: Apr 20, 2015
21. Apr 20, 2015

### goonking

oh wow, i meant to draw at 1.5 sec, not 2.5 sec, truly sorry!

22. Apr 20, 2015

### goonking

the answer should be 3 m/s then , right?

23. Apr 20, 2015

### AlephNumbers

Your value of A is not correct. It should be the maximum displacement. Also, make sure to use the correct value for ω.

24. Apr 20, 2015

### goonking

sorry, let me fix that.

A = 6 meters.
ω = π r/s

correct?

25. Apr 20, 2015

Looks good!