Solving for Initial Velocity v with No Oscillation

[Mark44]

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

How can v² be equal to a negative number?

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

 

[cloud360]

How can v² be equal to a negative number?

please can you tell me the answer r steps to get answer, i clearly don't know how to do this :(

 

[cloud360]

I think i got the correct solution to D, can someone tell me if my solution to D is correct or not

[PLAIN]http://img638.imageshack.us/img638/7313/2009b6d.gif

 

[Mark44]

Some of your work is correct, but you aren't reaching the right conclusions.
Your graph of V(x) is fine, and the min and max points are where you say they are. The local maximum for x = 3 is V(3) = 4/27, which is what you said.

What the graph of V(x) vs. x is telling you is that a particle must have a minumum energy of 4/27 so that doesn't oscillate. In part c you were supposed to show that if it started at x = 1 at a velocity of 1/2, it would oscillate between x = 2 and x = 3 - sqrt(5).
If you draw a horizontal line at V = 1/8 = .125 on your V(x) graph, this line intersects the potential energy curve at x = 2 and x = 3 - sqrt(5). It also intersects at x = 3 + sqrt(5). If the particle starts at x = 1 (with a potential energy of V(0) = 0, an initial velocity of 1/2 doesn't quite get it up over the hump whose height is 4/27 ~ 0.15. As a result, the particle is pulled back in the negative x direction to x = 1 but keeps going until it reaches a potential energy of .125 at x = 3 - sqrt(5). It stops momentarily, then changes direction, moving in the positive x direction, through x = 1 again, and on to x = 2. What's happening here is that since there wasn't enough velocity to give it enough energy, it keeps oscillating forever between x = 3 - sqrt(5) and x = 2.

For the last part, the particle starts at x = 1 (and V(1) = 0), and you need to find the velocity at which it should start to get it out of the "potential well".

As you have found, T + V >= 4/27. Since it's starting at x = 1, V(1) = 0, so you are solving T >= 4/27. I.e., .5v² >= 4/27.

To get up over the hump, the particle needs to be moving to the right, so v must be positive. You have been very cavalier about the signs of velocities. They do make a difference here, as a negative velocity indicates that the particle is moving to the right. What is the minimum positive velocity so that the particle keeps moving to the right, and doesn't oscillate between two x values?

All you have to do is solve the inequality .5v² >= 4/27.

 

[cloud360]

Some of your work is correct, but you aren't reaching the right conclusions.
Your graph of V(x) is fine, and the min and max points are where you say they are. The local maximum for x = 3 is V(3) = 4/27, which is what you said.

What the graph of V(x) vs. x is telling you is that a particle must have a minumum energy of 4/27 so that doesn't oscillate. In part c you were supposed to show that if it started at x = 1 at a velocity of 1/2, it would oscillate between x = 2 and x = 3 - sqrt(5).
If you draw a horizontal line at V = 1/8 = .125 on your V(x) graph, this line intersects the potential energy curve at x = 2 and x = 3 - sqrt(5). It also intersects at x = 3 + sqrt(5). If the particle starts at x = 1 (with a potential energy of V(0) = 0, an initial velocity of 1/2 doesn't quite get it up over the hump whose height is 4/27 ~ 0.15. As a result, the particle is pulled back in the negative x direction to x = 1 but keeps going until it reaches a potential energy of .125 at x = 3 - sqrt(5). It stops momentarily, then changes direction, moving in the positive x direction, through x = 1 again, and on to x = 2. What's happening here is that since there wasn't enough velocity to give it enough energy, it keeps oscillating forever between x = 3 - sqrt(5) and x = 2.

For the last part, the particle starts at x = 1 (and V(1) = 0), and you need to find the velocity at which it should start to get it out of the "potential well".

As you have found, T + V >= 4/27. Since it's starting at x = 1, V(1) = 0, so you are solving T >= 4/27. I.e., .5v² >= 4/27.

To get up over the hump, the particle needs to be moving to the right, so v must be positive. You have been very cavalier about the signs of velocities. They do make a difference here, as a negative velocity indicates that the particle is moving to the right. What is the minimum positive velocity so that the particle keeps moving to the right, and doesn't oscillate between two x values?

All you have to do is solve the inequality .5v² >= 4/27.

u are excellent :) thanks for all your time,

you said i have been very cavalier about the signs. so is it wrong to use the equals symbols? as opposed to the greater than and equal (>=) symbol? as we won't want the initialy velocity, its allowed to intersect once, so that's why i used equal symbol as it would be right at the edge.

i also know that i should have took the value from the V axis not x, so i should have tested

.5v² >= 4/27.
which gives v=sqrt (4/27)

and

.5v² >= 0
which gives v=0

so is the minimum velocity sqrt (4/27), if so, why is it not 0, as that is the other minimum point i got.

 

[Mark44]

u are excellent :) thanks for all your time,

You're welcome!

you said i have been very cavalier about the signs. so is it wrong to use the equals symbols?

By signs, I meant + and -, such as when you solved (x')² = 6, and got x' = +/- sqrt(6). Here we're not concerned with negative velocities (meaning that the particle is moving in the negative x direction.)

I wasn't referring to the fact that you are changing inequalities to equations, but now that you mention it, that is a concern as well.

If you are working with an inequality and end up with v² >= 9 (say), then v >= 3 or v <= - 3. If we are disallowing negative velocities, then we would end up with v >= 3, which means that the minimum velocity is 3.

as opposed to the greater than and equal (>=) symbol? as we won't want the initialy velocity, its allowed to intersect once, so that's why i used equal symbol as it would be right at the edge.

i also know that i should have took the value from the V axis not x, so i should have tested

.5v² >= 4/27.
which gives v=sqrt (4/27)

No, you need to multiply both sides by 2 first, so v² >= 8/27. You should end up here with an inequality, not an equation.

and

.5v² >= 0
which gives v=0

No, this is wrong. v can be any real number, positive or negative, and its square will always be >= 0.

More to the point, you only need to solve v² >= 8/27.

Practically speaking, if the particle starts from x = 1, with a velocity of 0, it's not going anywhere, so won't move at all, and hence doesn't oscillate. It definitely doesn't have enough energy to escape the potential well it's in. It has 0 energy - its kinetic energy (T) is 0 and its potential energy (V) is 0. In this problem, the energy can't be negative.

so is the minimum velocity sqrt (4/27), if so, why is it not 0, as that is the other minimum point i got.

 

[cloud360]

You're welcome!
By signs, I meant + and -, such as when you solved (x')² = 6, and got x' = +/- sqrt(6). Here we're not concerned with negative velocities (meaning that the particle is moving in the negative x direction.)

I wasn't referring to the fact that you are changing inequalities to equations, but now that you mention it, that is a concern as well.

If you are working with an inequality and end up with v² >= 9 (say), then v >= 3 or v <= - 3. If we are disallowing negative velocities, then we would end up with v >= 3, which means that the minimum velocity is 3.
No, you need to multiply both sides by 2 first, so v² >= 8/27. You should end up here with an inequality, not an equation.
No, this is wrong. v can be any real number, positive or negative, and its square will always be >= 0.

More to the point, you only need to solve v² >= 8/27.

Practically speaking, if the particle starts from x = 1, with a velocity of 0, it's not going anywhere, so won't move at all, and hence doesn't oscillate. It definitely doesn't have enough energy to escape the potential well it's in. It has 0 energy - its kinetic energy (T) is 0 and its potential energy (V) is 0. In this problem, the energy can't be negative.

q1) so is the answer v=sqrt(8/27) or v>=sqrt(8/27), as it is asking for minimum initial velocity, so i think i should get an equation, not in equality and say the minimum initial velocity is sqrt(8/27).

q2) also, i should always reject v=0 if i get it? right/ because it won't scape the potential well and won't move anywhere (if i thin of the graph as a hill, if i dropped a ball with v=0 at x=1 it won't move)

q3) i also have another similar question, caN you kindly tell me ifm y answer below for the question below is correct (so that i don't need to create another thread)

[PLAIN]http://img16.imageshack.us/img16/7806/2010b5mininitialvel.gif

 

[Mark44]

q1) so is the answer v=sqrt(8/27) or v>=sqrt(8/27), as it is asking for minimum initial velocity, so i think i should get an equation, not in equality and say the minimum initial velocity is sqrt(8/27).

The solution to the inequality (again assuming positive velocities) is v >= sqrt(8/27). So the answer to the question is "The minimum initial velocity is sqrt(8/27), or about .54." IOW, the answer is neither an equation nor an inequality, but a sentence that states what the smallest velocity is so that the particle keeps going. Also, you would normally include units for this velocity (m/sec, ft/sec, whatever), but I don't remember seeing any units given for distance, mass, or anything else.

q2) also, i should always reject v=0 if i get it? right/ because it won't scape the potential well and won't move anywhere (if i thin of the graph as a hill, if i dropped a ball with v=0 at x=1 it won't move)

That's probably a good idea, but try to get more of a feel for what things mean in these problems, and that will help you eliminate algebraic solutions that don't have any meaning in a physics-type problem.

q3) i also have another similar question, caN you kindly tell me ifm y answer below for the question below is correct (so that i don't need to create another thread)

[PLAIN]http://img16.imageshack.us/img16/7806/2010b5mininitialvel.gif[/QUOTE]

I don't have time to look at it now, but I will do so later.

 

[Mark44]

That's what I get, but you can say it a bit cleaner.
If the particle starts at x = 1, its potential energy is 1, as you found. I.e., V(1) = 1.

At x = 2, the particle's potential energy is V(2) = 3 ln(2) - 1.

In order for the particle to get over the hump (IOW, not oscillate between two points to the left and right of x = 1), it must have a total energy >= 3 ln(2) - 1

Since its potential energy at x = 1 is V(1) = 1, it needs at least V(2) - V(1) more energy to get out of the potential well, and that energy can come only from kinetic energy T.

So T >= V(2) - V(1)
==> .5v² >= 3 ln(2) - 1 - 1
==> v² >= 6 ln(2) - 4
==> v >= sqrt( 6 ln(2) - 4)

That is, the minimum velocity for no oscillation is sqrt(6 ln(2) - 4).