Where is the Velocity of a Particle Moving Along the x-axis a Maximum?

[interxavier]

Homework Statement

A particle can only move along the x axis. Forces act on it so that its potential energy function is U(x) = 1/2*k1*x^2 + 1/4*k2*x^4 where k1 and k2 are positive. The particle is started at x = a with zero velocity.
a.) Where is the velocity a maximum? What is its magnitude?
b.) Where else will the velocity be zero?
c.) What is the force on the particle as a function of x?

Homework Equations

F = d/dx -U(x)
W = -U(x2) + U(x1)

The Attempt at a Solution

It's asking for the point where the velocity is maximum so the acceleration has to equal zero or not exist. The function exists at all x-values so we need to find the acceleration as a function of x.

since F = d/dx -U(x), F = -(k1*x + k2*x^3)

F = ma

ma = -(k1*x + k2*x^3)
a = -1/m*(k1*x + k2*x^3)
When a = 0
0 = -1/m*(k1*x + k2*x^3)
-k1*x = k2*x^3
x^2 = -k1/k2
x = sqrt(-k1/k2)

I'm completely lost at this point. Thanks for the help!

 

[kuruman]

You can't say x=sqrt(-k1/k2) for two good reasons (a) x is an independent variable - it tells you where the particle is - so it cannot be constant and (b) the negative of a square root has no physical meaning whereas x denotes where the particle is at a given time.

Look at your expression again,

0 = -1/m*(k1*x + k2*x^3)

For what value(s) of x is the right side zero?

For part (b): You know that at x = a the kinetic energy is zero and the potential energy is
U = (1/2)k₁a²+(1/4)k₂a⁴. Can you find another value for x, other x = +a where the potential energy has the same value? If you can, then you know that by conservation of energy the kinetic energy will be zero there as well.