Solving for Initial Velocity v with No Oscillation

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Homework Help Overview

The discussion revolves around determining an initial velocity \( v \) that results in no oscillation for a particle described by a potential energy function. Participants are examining the conditions under which the particle's motion can be analyzed without oscillatory behavior, referencing specific equations and graphical representations of energy and potential.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the implications of initial conditions on the particle's motion, questioning the treatment of position and velocity as constants. They explore the relationship between kinetic and potential energy and seek to clarify the conditions that lead to non-oscillatory motion.

Discussion Status

There is an active exploration of the mathematical relationships involved, with participants providing insights into the energy equation and its implications for velocity. Some guidance has been offered regarding the interpretation of energy levels and their relation to oscillation, but no consensus has been reached on the specific values or intervals that ensure no oscillation.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available for discussion. The original poster has shared their attempts at solutions, and there is a focus on verifying these attempts against the established equations and principles of physics.

cloud360
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?!??Find an initial velocity v such that there's no oscillation,is my solution correct

Homework Statement



Can someone kindly tell me if my solution to question C and D is correct, if not, what did i do wrong?

[PLAIN]http://img9.imageshack.us/img9/4914/b52009cd.gif

Homework Equations


The Attempt at a Solution



This is my solution to C

[PLAIN]http://img51.imageshack.us/img51/540/solc.gif

This is my solution to d

[PLAIN]http://img35.imageshack.us/img35/4426/solvol.gif

for reference, this is the sketch of V(x) with E=0.125 superimposed

[PLAIN]http://img220.imageshack.us/img220/213/solbwe.gif
 
Last edited by a moderator:
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anyone?
 


Mark44 said:
Did you do part a?

yes i did, do u need solution to part a also?
 


If you've done part a, then you know that (1/2)mx'2 = E0 - 1/x + 2/x2 - 1/x3.

If the particle oscillates between the two given x values, it must change direction, and when it does so, x' = 0. For what values of x does the equation above make x' = 0?

In your work, it seems that you treated x and x' as constants, which is not valid.
 


Mark44 said:
If you've done part a, then you know that (1/2)mx'2 = E0 - 1/x + 2/x2 - 1/x3.

If the particle oscillates between the two given x values, it must change direction, and when it does so, x' = 0. For what values of x does the equation above make x' = 0?

In your work, it seems that you treated x and x' as constants, which is not valid.

i did that because it said x=1 and x'=0.5
 


cloud360 said:
i did that because it said x=1 and x'=0.5
It doesn't actually say that. What it says is that the initial position is 1 and the initial velocity is .5. These are the position and time at t = 0. Obviously (I hope it's obvious), the position and velocity are not constant, so you can't treat them as if they were constant.
 


Mark44 said:
It doesn't actually say that. What it says is that the initial position is 1 and the initial velocity is .5. These are the position and time at t = 0. Obviously (I hope it's obvious), the position and velocity are not constant, so you can't treat them as if they were constant.

this is what my professor did, he treated them as constant. here is an example question and answer (where he treats as constants, he finds E, then subs back in)

(in these questions m=1)
[PLAIN]http://img708.imageshack.us/img708/14/energyequation.gif
 
Last edited by a moderator:


Actually, you were on the right track in your first post, but I didn't catch what you were doing. There is an error, though, and that threw me off.

Starting from this equation:
1/x - 2/x^2 - 1/x^3 = .125

Multiplying by x^3 is what you want to do, but you get
x^2 - 2x + 1 = .125x^3

This is equivalent to x^3 - 8x^2 + 16x - 8 = 0.

This equation can be factored into a linear factor (one factor is x - 2) and a quadratic.
 
  • #10


Mark44 said:
Actually, you were on the right track in your first post, but I didn't catch what you were doing. There is an error, though, and that threw me off.

Starting from this equation:
1/x - 2/x^2 - 1/x^3 = .125

Multiplying by x^3 is what you want to do, but you get
x^2 - 2x + 1 = .125x^3

This is equivalent to x^3 - 8x^2 + 16x - 8 = 0.

This equation can be factored into a linear factor (one factor is x - 2) and a quadratic.

using wolfram

http://www.wolframalpha.com/input/?i=x^2-2x%2B1-0.125x^3

i get solutions

x=2
x=0.76 (3-sqrt5)

x=5.236

now i am confused, how do between which values it oscilaltes, does it oscillate between min and max values i.e .75 and 5.236 or 2 and 5.236
 
  • #11


also, can you kindly confirm if my solution to d is correct
 
  • #12


Did you graph y = x^3 - 8x^2 + 16x - 8?

What you have found are the three x-intercepts of this function. At each of these three values, x' = 0, which means that V = E0. At all other positions x, the value of x^3 - 8x^2 + 16x - 8 is either positive or negative. Tie this information into your equations for T and V and see what that says about the particle's motion.
 
  • #13


Mark44 said:
Did you graph y = x^3 - 8x^2 + 16x - 8?

What you have found are the three x-intercepts of this function. At each of these three values, x' = 0, which means that V = E0. At all other positions x, the value of x^3 - 8x^2 + 16x - 8 is either positive or negative. Tie this information into your equations for T and V and see what that says about the particle's motion.

ok

plugging x=2 into T+V ==> E=1/8=0.125

x=0.76 (3-sqrt5) into T+V==> E=1/8=0.125

x=5.236 into T+V==>E=1/8 (approximately)

does that mean them minimum initial velocity is 0 because all values of E and x, gives 0 when we differentiate
 
  • #14


cloud360 said:
ok

plugging x=2 into T+V ==> E=1/8=0.125

x=0.76 (3-sqrt5) into T+V==> E=1/8=0.125

x=5.236 into T+V==>E=1/8 (approximately)

does that mean them minimum initial velocity is 0 because all values of E and x, gives 0 when we differentiate
I don't understand why you're doing this. At the three roots, T = 0 because x' = 0, and these three values of x are the ones for which V = E0 = .125.

If I'm remembering my physics, the equation T + V = E0 is saying that kinetic energy (T) + potential energy (V) is constant (E0).

At the x values x = 2, x = 3 +/- sqrt(5), x' = 0. At all other positions (values of x), x' is NOT zero, which means that the velocity must be either positive (particle moving to the right on the x-axis) or negative (particle moving to the left on the x-axis).
 
  • #15


Mark44 said:
I don't understand why you're doing this. At the three roots, T = 0 because x' = 0, and these three values of x are the ones for which V = E0 = .125.

If I'm remembering my physics, the equation T + V = E0 is saying that kinetic energy (T) + potential energy (V) is constant (E0).

At the x values x = 2, x = 3 +/- sqrt(5), x' = 0. At all other positions (values of x), x' is NOT zero, which means that the velocity must be either positive (particle moving to the right on the x-axis) or negative (particle moving to the left on the x-axis).

so how do i found this velocity? please can you give me a hint (i read your previous message and i am still confused as to what i must do)
 
  • #16


Your energy equation works out to
(1/2)mv2 + 1/x - 2/x2 + 1/x3 = 1/8, with v = x'

Solving for v2, we get
v2 = -2/m(1/x - 2/x2 + 1/x3 + 1/8)

When v = 0, you have -2/m(1/x - 2/x2 + 1/x3 + 1/8) = 0, and you have already found the x values for which v = 0 (x = 2, x = 3 +/- sqrt(5)).

Now v2 is necessarily >= 0, so 1/x - 2/x2 + 1/x3 + 1/8 must be <= 0, since it's being multiplied by -2/m, a negative constant.



For what intervals is the graph of y = 1/x - 2/x2 + 1/x3 + 1/8 strictly negative? It has to be one or more of these intervals: (-inf, 3 - sqrt(5)), (3 - sqrt(5), 2), (2, 3 + sqrt(5)), or ( 3 + sqrt(5), inf).
 
  • #17


Mark44 said:
Your energy equation works out to
(1/2)mv2 + 1/x - 2/x2 + 1/x3 = 1/8, with v = x'

Solving for v2, we get
v2 = -2/m(1/x - 2/x2 + 1/x3 + 1/8)

When v = 0, you have -2/m(1/x - 2/x2 + 1/x3 + 1/8) = 0, and you have already found the x values for which v = 0 (x = 2, x = 3 +/- sqrt(5)).

Now v2 is necessarily >= 0, so 1/x - 2/x2 + 1/x3 + 1/8 must be <= 0, since it's being multiplied by -2/m, a negative constant.
For what intervals is the graph of y = 1/x - 2/x2 + 1/x3 + 1/8 strictly negative? It has to be one or more of these intervals: (-inf, 3 - sqrt(5)), (3 - sqrt(5), 2), (2, 3 + sqrt(5)), or ( 3 + sqrt(5), inf).
did u get those intervals from the solutions we found (x = 2, x = 3 +/- sqrt(5)). How do i know what the minimum velocity is from those intervals, such that there is no oscillation?

also, if there is no oscillation, should we choose values which would give a value of E such that it does not intersect V(x) twice, i am not sure why u are doing what your doing.

if i just look at graph, i can see if its greater than the maximum i.e E>0 , then there is no oscillation (since we would have only 1 intersection)

or if where less than -7 i.e E<-7 (the minimum), sorry in my graph i did not label the negative y
 
  • #18


cloud360 said:
did u get those intervals from the solutions we found (x = 2, x = 3 +/- sqrt(5)).
Yes, of course. Before going on to the other questions you asked, how about answering my question about the interval(s) to use?
cloud360 said:
How do i know what the minimum velocity is from those intervals, such that there is no oscillation?

also, if there is no oscillation, should we choose values which would give a value of E such that it does not intersect V(x) twice, i am not sure why u are doing what your doing.

if i just look at graph, i can see if its greater than the maximum i.e E>0 , then there is no oscillation (since we would have only 1 intersection)

or if where less than -7 i.e E<-7 (the minimum), sorry in my graph i did not label the negative y
 
  • #19


You have a graph that you say is V(x) in your first post. I don't know what you graphed, but the potential, V(x), doesn't look anything like what you have.

Here's a graph of V(x) = 1/x - 2/x^2 + 1/x^3 from wolframalpha:
http://www.wolframalpha.com/input/?i=Plot[x^(-3)+-+2/x^2+++x^(-1),+{x,+0,+8}]

From this graph, it appears that V(1) = 0, and that is confirmed if you use the formula above. Also, for x > 1, V(x) <= ~ .15. Checking some more, I found that V(2) = 1/8 and that V(3 - sqrt(5)) = 1/8 (exactly). Those two numbers should look familiar.

What happens to x' (= v) when x is between 3 - sqrt(5) and 2?
 
  • #20


we have 3 points, 3 - sqrt(5) and 2 and 3 + sqrt(5), and we proved it oscillates between 2 and (3-sqrt5), doesn't it also oscillate between 3-sqrt5(smallest value) and 3+sqrt5(highest value)?

this is something i really want to clear up !
------------
Anyway, to answer your question

Mark44 said:
What happens to x' (= v) when x is between 3 - sqrt(5) and 2?

x'=0 and it oscillates in between those 2 points? meaning outside those 2 points it doesn't oscillate, also this goes back to my question, doesn't is also osccillate between 3-sqrt5(smallest value of 3) and 3+sqrt5(highest value)?

at x<=0 and x>=0.15, it does not osccilate !

do we then sub x=0 and x=0.15 back into T+V and try find r'=v?
 
  • #21


cloud360 said:
we have 3 points, 3 - sqrt(5) and 2 and 3 + sqrt(5), and we proved it oscillates between 2 and (3-sqrt5), doesn't it also oscillate between 3-sqrt5(smallest value) and 3+sqrt5(highest value)?
I don't recall seeing any of your work that proves that the particle oscillates between 2 and 3 - sqrt(5).
cloud360 said:
this is something i really want to clear up !
------------
Anyway, to answer your question



x'=0 and it oscillates in between those 2 points? meaning outside those 2 points it doesn't oscillate, also this goes back to my question, doesn't is also osccillate between 3-sqrt5(smallest value of 3) and 3+sqrt5(highest value)?
The question was
What happens to x' (= v) when x is between 3 - sqrt(5) and 2?
No, x' is not 0 between those two numbers. x' = 0 when x = 3 - sqrt(5) and x' = 0 when x = 2, but in between x' is not zero. The only values of x for which x' = 0 are 3 - sqrt(5), 2, and 3 + sqrt(5). The only places x' (or v) can possibly change signes are at these three numbers, so within any of the intervals (3 - sqrt(5), 2), (2, 3 + sqrt(5)), v must be one sign throughout.

When the particle is at exactly x = 3 - sqrt(5), x' = v = 0. What's happening to the particle if v = 0?

Also, at exactly x = 2, x' = v = 0. Same question as above.

For 3 - sqrt(5) < x < 2, what sign is v?
For 2 < x < 3 + sqrt(5), what sign is v?


cloud360 said:
at x<=0 and x>=0.15, it does not osccilate !
How can x possibly be negative? And where 0.15 come from?
cloud360 said:
do we then sub x=0 and x=0.15 back into T+V and try find r'=v?
 
  • #22


Mark44 said:
I don't recall seeing any of your work that proves that the particle oscillates between 2 and 3 - sqrt(5).
The question was

No, x' is not 0 between those two numbers. x' = 0 when x = 3 - sqrt(5) and x' = 0 when x = 2, but in between x' is not zero. The only values of x for which x' = 0 are 3 - sqrt(5), 2, and 3 + sqrt(5). The only places x' (or v) can possibly change signes are at these three numbers, so within any of the intervals (3 - sqrt(5), 2), (2, 3 + sqrt(5)), v must be one sign throughout.

When the particle is at exactly x = 3 - sqrt(5), x' = v = 0. What's happening to the particle if v = 0?

Also, at exactly x = 2, x' = v = 0. Same question as above.

For 3 - sqrt(5) < x < 2, what sign is v?
For 2 < x < 3 + sqrt(5), what sign is v?


How can x possibly be negative? And where 0.15 come from?

i got x=0.15 from this graph

http://www.wolframalpha.com/input/?i=Plot[x^(-3)+-+2/x^2+++x^(-1),+{x,+0,+8}]

(i really appreciate that you took your time to help out again)

my professor said if it doesn't intersect V(x) twice then it does not osccilate
 

Attachments

  • poscil.gif
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  • #24


cloud360 said:
my professor said if it doesn't intersect V(x) twice then it does not osccilate

Try writing what your professor said without using "it".
 
  • #25


Mark44 said:
Try writing what your professor said without using "it".

when you say without using "it", i am assuming that you thought i knew the answer and you are referring to "it" as the solution?

the attachment that i added was the solution to another question, not this one (it explains that it needs to intersect twice for oscillation, but i don't know how to use that fact). anyway back to my question.

Code: 
For 3 - sqrt(5) < x < 2, what sign is v?
For 2 < x < 3 + sqrt(5), what sign is v?

for 3 - sqrt(5) < x < 2, v is always positive

for 2 < x < 3 + sqrt(5) its also always positive from the graph of V(x)

http://www.wolframalpha.com/input/?i=Plot%5Bx^%28-3%29+-+2%2Fx^2+%2B+x^%28-1%29%2C+{x%2C+0%2C+8}%5D

anyway the smallest v for which there is no oscillation is v=0 or v=0.15

right? it has to be one of those 2? at those points it does not intersect twice, 1 is a maximum the other is a minimum, and either side of it, we won't have 2 intersections
 
  • #26


cloud360 said:
when you say without using "it", i am assuming that you thought i knew the answer and you are referring to "it" as the solution?
No, I don't know what you meant by "it" and I would like you to tell me what you meant in a sentence that doesn't use "it."
cloud360 said:
the attachment that i added was the solution to another question, not this one (it explains that it needs to intersect twice for oscillation, but i don't know how to use that fact). anyway back to my question.

Code: 
For 3 - sqrt(5) < x < 2, what sign is v?
For 2 < x < 3 + sqrt(5), what sign is v?

for 3 - sqrt(5) < x < 2, v is always positive

for 2 < x < 3 + sqrt(5) its also always positive from the graph of V(x)

http://www.wolframalpha.com/input/?i=Plot%5Bx^%28-3%29+-+2%2Fx^2+%2B+x^%28-1%29%2C+{x%2C+0%2C+8}%5D

anyway the smallest v for which there is no oscillation is v=0 or v=0.15

right? it has to be one of those 2? at those points it does not intersect twice, 1 is a maximum the other is a minimum, and either side of it, we won't have 2 intersections
This is the graph of V(x), not v. V is the potential energy at a distance x. I'm asking you about v, which is velocity. I've been very careful to note that v means x'.

If V = 0, a horizontal line won't cut the graph at all. If V >= .15, a horizontal line cuts the graph only once.

I'm getting somewhat discouraged, since you seem to have such a hard time taking even the smallest steps on your own. It might help if you went back through this whole thread and reread it carefully. If you have a specific question about anything that has already appeared, ask it.
 
  • #27


Mark44 said:
Did you graph y = x^3 - 8x^2 + 16x - 8?

What you have found are the three x-intercepts of this function. At each of these three values, x' = 0, which means that V = E0. At all other positions x, the value of x^3 - 8x^2 + 16x - 8 is either positive or negative. Tie this information into your equations for T and V and see what that says about the particle's motion.

Mark44 said:
I don't understand why you're doing this. At the three roots, T = 0 because x' = 0, and these three values of x are the ones for which V = E0 = .125.

If I'm remembering my physics, the equation T + V = E0 is saying that kinetic energy (T) + potential energy (V) is constant (E0).

At the x values x = 2, x = 3 +/- sqrt(5), x' = 0. At all other positions (values of x), x' is NOT zero, which means that the velocity must be either positive (particle moving to the right on the x-axis) or negative (particle moving to the left on the x-axis).

Mark44 said:
Your energy equation works out to
(1/2)mv2 + 1/x - 2/x2 + 1/x3 = 1/8, with v = x'

Solving for v2, we get
v2 = -2/m(1/x - 2/x2 + 1/x3 + 1/8)

When v = 0, you have -2/m(1/x - 2/x2 + 1/x3 + 1/8) = 0, and you have already found the x values for which v = 0 (x = 2, x = 3 +/- sqrt(5)).

Now v2 is necessarily >= 0, so 1/x - 2/x2 + 1/x3 + 1/8 must be <= 0, since it's being multiplied by -2/m, a negative constant.
For what intervals is the graph of y = 1/x - 2/x2 + 1/x3 + 1/8 strictly negative? It has to be one or more of these intervals: (-inf, 3 - sqrt(5)), (3 - sqrt(5), 2), (2, 3 + sqrt(5)), or ( 3 + sqrt(5), inf).

ok you said at the oscillating points x'=0, so we would only have the potential part V when it oscillates.

now i have to assume that x' is not 0, to get my answers, so let x'=v

v2 = -2/m(1/x - 2/x2 + 1/x3 + 1/8)

http://www.wolframalpha.com/input/?i=v2+=+-2/1(1/x+-+2/x2+++1/x3+++1/8),++x=1

What happens to x' (= v) when x is between 3 - sqrt(5) and 2?

x'=v is an imaginary number?

example when x=0.83

http://www.wolframalpha.com/input/?i=v2+=+-2/1(1/x+-+2/x2+++1/x3+++1/8),++x=3+-+sqrt(5)+0.1

we get v= +/- 0.5i, if we assume x=1, which is also between that interval
 
  • #28


cloud360 said:
ok you said at the oscillating points x'=0, so we would only have the potential part V when it oscillates.

now i have to assume that x' is not 0, to get my answers, so let x'=v

v2 = -2/m(1/x - 2/x2 + 1/x3 + 1/8)

http://www.wolframalpha.com/input/?i=v2+=+-2/1(1/x+-+2/x2+++1/x3+++1/8),++x=1
This graph is not very useful, as it has x on the vertical axis and v on the horizontal axis. A more useful graph would be y = -2/1 (1/x - 2/x2 + 1/x3). What you're looking for are the x intervals for which y > 0.
cloud360 said:
What happens to x' (= v) when x is between 3 - sqrt(5) and 2?

x'=v is an imaginary number?
How does this make any sense at all? v is the velocity of a physical particle. Is it meaningful to look at imaginary or complex values of velocity?
cloud360 said:
example when x=0.83

http://www.wolframalpha.com/input/?i=v2+=+-2/1(1/x+-+2/x2+++1/x3+++1/8),++x=3+-+sqrt(5)+0.1

we get v= +/- 0.5i, if we assume x=1, which is also between that interval
 
  • #29


Mark44 said:
This graph is not very useful, as it has x on the vertical axis and v on the horizontal axis. A more useful graph would be y = -2/1 (1/x - 2/x2 + 1/x3). What you're looking for are the x intervals for which y > 0.
How does this make any sense at all? v is the velocity of a physical particle. Is it meaningful to look at imaginary or complex values of velocity?

this part (1/x - 2/x2 + 1/x3) represents V, and i can see from this graph

http://www.wolframalpha.com/input/?i=Plot%5Bx^%28-3%29+-+2%2Fx^2+%2B+x^%28-1%29%2C+{x%2C+0%2C+8}%5D

which u said is not that helpful, i can see when V>0.15, there is no osccilation

So is we say 0.15=(1/x - 2/x2 + 1/x3)

then -2*0.15=-0.3 = v^2

does this mean the minimum initial velocity is +/- sqrt3
 
  • #30


anyone?
 

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