Conservation of Energy Problem

[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.

 

[tomcue]

Your approach is on the right track. To find the maximum velocity, we need to find the point where the acceleration is equal to zero. This occurs when the net force on the particle is equal to zero. So, we can set the force equation equal to zero and solve for x.

F = -(k1*x + k2*x^3) = 0
x(k1 + k2*x^2) = 0
x = 0 or x = ±√(-k1/k2)

Since we are looking for the maximum velocity, we can ignore the solution x = 0 (as this would give us zero velocity). So, the maximum velocity occurs at x = ±√(-k1/k2). To find the magnitude of the velocity, we can use the equation v = √(2/m * (E - U(x))), where E is the total energy of the particle.

a.) So, at x = ±√(-k1/k2), the velocity will be maximum and its magnitude will be v = √(2/m * (E - U(x))) = √(2/m * (E - (1/2*k1*√(-k1/k2)^2 + 1/4*k2*√(-k1/k2)^4))) = √(2/m * (E - k1*(-k1/k2) - k2*(-k1/k2)^2)) = √(2/m * (E + k1^2/k2 + k1^2/k2^2))

b.) The velocity will be zero at x = a (initial position) and at the point where the particle reaches its maximum displacement, x = ±√(-k1/k2).

c.) The force on the particle as a function of x is given by F = d/dx (-U(x)) = -(k1*x + k2*x^3). This is the same equation you found, but it is important to note that this is the net force on the particle, as it is the derivative of the potential energy function with respect to x.