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Displacement, velocity and acceleration

  • Thread starter lha08
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  • #1
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Homework Statement


I'm having a lot of trouble trying to figure out the correlation between the displacement, velocity and acceleration and when they are positive or negative according to the other. For example, when displacement is positive, the velocity is negative and in another question, when a particular displacement was negative, the velocity was either positive or negative..why is that exactly? One other question, does the velocity ever decrease or is the negative just indicating the direction?
Thanks.


Homework Equations





The Attempt at a Solution

 
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Answers and Replies

  • #2
Born2bwire
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A good physical example of a simple harmonic oscillator (SHO) is a weight on a compressible spring. When you pull the weight, the spring stretches and exerts a force back towards the equilibrium position. When you push the weight back past the position of equilibrium, the spring will again exert an opposite force pushing back towards equilibrium.

So you can see just from that simple example that if the equilibrium position of the SHO is x=0, then when the mass is at some point x>0, then there must be a restoring force back towards x=0, which would be in the -x direction. The opposite is true for when the mass is at some point x<0.

However, just because the force is negative does not mean the velocity is negative. If the mass is oscillating, it will move past the point of equilibrium, start to slow down, stop, and then start accelerating in the opposite direction. From the point that it stops at some position x<0 to the point it stops at some position x>0, the velocity is positive. The velocity only changes sign when the mass stops and changes direction.

The equations for a spring (along one dimension) are:
[tex]F = ma = m\frac{d v}{dt} = -kx[/tex]
[tex]v = \frac{dx}{dt}[/tex]

You can use calculus to derive the expressions for the position as a function of time, velocity as a function of time, and then acceleration as a function of time and this will confirm mathematically the results of any thought exercise.
 
  • #3
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Let us take displacement as any periodic function of time{The basic condition for S.H.M.}
suppose, x{displacement}=a{amplitude}sin{the simplest periodic function}(wt){phase of the motion}
[itex]v=\dot{x}=w\sqrt{a^2-x^2} [/itex]
add a plus minus symbol infront of w[dunno how to do that in latex] and the equation will answer all your questions
 
  • #4
Born2bwire
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Let us take displacement as any periodic function of time{The basic condition for S.H.M.}
suppose, x{displacement}=a{amplitude}sin{the simplest periodic function}(wt){phase of the motion}
[itex]v=\dot{x}=w\sqrt{a^2-x^2} [/itex]
add a plus minus symbol infront of w[dunno how to do that in latex] and the equation will answer all your questions
\pm = [tex]\pm[/tex]

But if the displacement is sinusoidal, then the velocity, it's time derivative, will also be sinusoidal.

[tex]v = a\omega\cos(\omega t)[/tex]
 
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  • #5
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[tex]v = -a\omega\cos(\omega t)[/tex]
[How did you get that minus sign?]
plug in x using the simplest of trignometric identities
and you will get [itex]v=\dot{x}=\pm w \sqrt{a^2-x^2} [/itex]{This is also sinusoidal}
 
  • #6
Born2bwire
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[tex]v = -a\omega\cos(\omega t)[/tex]
[How did you get that minus sign?]
plug in x using the simplest of trignometric identities
and you will get [itex]v=\dot{x}=\pm w \sqrt{a^2-x^2} [/itex]{This is also sinusoidal}
Oh, I see what you did. Yeah, the minus sign shouldn't be there.
 
  • #7
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However, just because the force is negative does not mean the velocity is negative. If the mass is oscillating, it will move past the point of equilibrium, start to slow down, stop, and then start accelerating in the opposite direction. From the point that it stops at some position x<0 to the point it stops at some position x>0, the velocity is positive. The velocity only changes sign when the mass stops and changes direction.
Like in my textbook, they relate the displacement with the velocity and it shows that when the displacement is positive until it reaches the equilibrium point, the slope on the velocity graph is decreasing from the positive to negative y-axis and likewise for when the displacement on the graph is in the negative region, the slope of the velocity is increasing from the negative to positive y-axis...it the velocity positive or negative then?
 
  • #8
Born2bwire
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Acceleration would is related to the change in velocity over time. So if the velocity is decreasing, then it is the acceleration that is negative (which is in the same direction as the force). As long as the velocity > 0 on the graph, then it is positive. So do not look at how the velocity is changing to find whether or not it is positive, just look at the value of the velocity itself.
 
  • #9
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Let us take displacement as any periodic function of time{The basic condition for S.H.M.}
suppose, x{displacement}=a{amplitude}sin{the simplest periodic function}(wt){phase of the motion}
[itex]v=\dot{x}=w\sqrt{a^2-x^2} [/itex]
add a plus minus symbol infront of w[dunno how to do that in latex] and the equation will answer all your questions
You lost me..is that formula supposed to represent the formula for the displacement of the SHM with just a plus and minus sign added onto it?
 
  • #10
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I don't know what level you are at lha, but I'm thinking all of the equations above might be confusing you.

In my opinion, a simpler model for a SHM oscillator is a simple pendulum.

Imagine it was swinging in the plane of the page. And take right as the positive direction.

At the left and right most points of the swing, the velocity is zero (it stops at these points.)

In the middle, the velocity is maximum.

So on it's way to the middle, it is speeding up. Once it's past the middle it is slowing down. Obviously you can talk about the direction as well - if it is moving to the right then it is positive.

You can think about the acceleration from this last point....
 
  • #11
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Acceleration would is related to the change in velocity over time. So if the velocity is decreasing, then it is the acceleration that is negative (which is in the same direction as the force). As long as the velocity > 0 on the graph, then it is positive. So do not look at how the velocity is changing to find whether or not it is positive, just look at the value of the velocity itself.
Just to make sure i understand:
1. If an object starts at its maximum amplitude, the velocity at that point is zero?
2. As it heads towards the equilibrium point, the velocity is negative and increasing?
3. As it passes the equilibrium point, it is negative and decreasing?
4. Eventually it stops at it max amplitude, and the velocity becomes positive and increasing?
5. When it passes the equilibrium point, the velocity is till positive but decreasing?

Is this right??
 
  • #12
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You lost me..is that formula supposed to represent the formula for the displacement of the SHM with just a plus and minus sign added onto it?
That formula represents velocity as a function of displacement.
 
  • #13
Born2bwire
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Just to make sure i understand:
1. If an object starts at its maximum amplitude, the velocity at that point is zero?
2. As it heads towards the equilibrium point, the velocity is negative and increasing?
3. As it passes the equilibrium point, it is negative and decreasing?
4. Eventually it stops at it max amplitude, and the velocity becomes positive and increasing?
5. When it passes the equilibrium point, the velocity is till positive but decreasing?

Is this right??
Yep, the only thing I would add is that in step one, the object is starting at the maximum amplitude in the positive displacement axis (x>0) and in step 4 the object is stopping at the maximum amplitude in the negative displacemen (x<0).
 

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