Solving for Initial Velocity v with No Oscillation

[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).