- #36
Mark44
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You're welcome!cloud360 said:u are excellent :) thanks for all your time,
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.)cloud360 said:you said i have been very cavalier about the signs. so is it wrong to use the equals symbols?
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.cloud360 said: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, this is wrong. v can be any real number, positive or negative, and its square will always be >= 0.cloud360 said:and
.5v2 >= 0
which gives v=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.
cloud360 said:so is the minimum velocity sqrt (4/27), if so, why is it not 0, as that is the other minimum point i got.