# Velocity from graph question

1. Feb 14, 2010

### optoracko

1. The problem statement, all variables and given/known data
Given a position time graph and velocity time graph for 1 cycle of an oscillating pendulum, if the curve for velocity passes through 0 m/s, can you say that velocity at that point is zero, even though there are no times in the position-time graph where the graph is a straight line?

2. Relevant equations
Velocity = displacement / time
Instantaneous velocity = slope of tangent on position time graph = displacement / time

3. The attempt at a solution
This is just a general physics question. Basically the graph for the pos - time graph is a bell curve and velocity time graph is a basic sine function.

I was thinking that I could the INSTANTANEOUS velocity is zero at that point, but not the average. I'm not sure if that is correct, but it makes sense since it is at a certain point in time.

Since the position time graph is a bell curve, could I say that relevant to the starting position, since the finishing position is at the starting, that there is no displacement and therefore velocity is 0, even though the position time graph is a curve and not a horizontal line?

2. Feb 14, 2010

### pgardn

Is your position v. time graph in 1-D? For example if the pendulum is swinging at a small enough angle if you look at the position of the pendulum with respect to the horizontal you should also see a sine or cosine graph depending on where you want to start from (really interchagebable). Ahhh I see it is one cycle so it should be a cosine graph.

If you then put the velocity in that reference (x-velocity) you should see exactly the opposite type of graph. Where position is at its max, the velocity is at its minimum. Have you looked at the graphs one right over the other thru time?

And there are times in a positon v. time graph where the slope is zero as inst. velocity is the slope of the position v. time graph no?

3. Feb 14, 2010

### pgardn

Is your position v. time graph in 1-D? For example if the pendulum is swinging at a small enough angle if you look at the position of the pendulum with respect to the horizontal you should also see a sine or cosine graph depending on where you want to start from (really interchagebable). Ahhh I see it is one cycle so it should be a cosine graph. A bell curve is just desciptive of the general shape as it is really a sine or in your case, a cosine graph?

If you then put the velocity in that reference (x-velocity) you should see exactly the opposite type of graph. Where position is at its max, the velocity is at its minimum. Have you looked at the graphs one right over the other thru time?

And there are times in a positon v. time graph where the slope is zero as inst. velocity is the slope of the position v. time graph no?

Also, how do you figure out the avg. velocity if the acceleration is clearly changing? It must be if the velocity v. time graph is curved.

4. Feb 14, 2010

### optoracko

http://img704.yfrog.com/img704/3217/asfnasf.jpg [Broken]

I'm sorry, I'm not really understanding what is being hinted at. The velocity is at it's minimum (zero, magnitude-wise) at the max position. For this position time graph, it appears as if there are no times where the slope is zero.

My wording may have been unclear, apologies if so.

Last edited by a moderator: May 4, 2017
5. Feb 14, 2010

### pgardn

Look at the 0.6s mark on the position v. time graph. This represents maximum displacement but this is also when the object is changing direction. So if the object swings up and then gets ready to swing back down it must come to an instantaneous halt v= 0 m/s at this time. The slope is zero at the 0.6s mark on the position v. time graph, yes?

Last edited by a moderator: May 4, 2017
6. Feb 14, 2010

### optoracko

Ah, which is sort of related to acceleration due to gravity.

Oh I see. I thought it only applied when moving just up and down, ex. jumping. Does this mean that anything moving "up" would experience this? Even if the angle at which it was moving was very low?

7. Feb 14, 2010

### pgardn

Experience what, changing direction at v= 0 m/s? If you mean a ball thrown up and coming to its highest point yes, but in this case the ball is always has a constant acceleration down, which is not true in the case of the pendulum.

And the position v. time graph for a ball thrown up in the vertical would be parabolic, not sinusoidal. The velocity v. time graph for the ball would be a straight (constant slope/constant accel.) diagonal line down.

Last edited: Feb 14, 2010