Solve Oscillator (not SHM) Homework: Potential Energies & Time Periods

[iloveannaw]

Homework Statement

For a certain oscillator the net force on the body with mass {m} is given by F_x = - cx^3.
a) What is the potential energy function for this oscillator if we take U = 0 at x = 0?
b) One-quarter of a period is the time for the body to move from x = 0 to x = A. Calculate this time and hence the period.

Homework Equations

a) $$U = \int F dx$$
b) $$E = \frac{1}{2}(mv^{2} + kx^{2})$$

$$T = \int\frac{dx}{v(x)}$$

The Attempt at a Solution

ok, the first part is pretty straight forward. just integrating w.r.t x yields $$\frac{1}{4}cx^{4}$$

$$E = constant = \frac{1}{4}cA^{4} = \frac{1}{2}mv^{2} + \frac{1}{4}cx^{4}$$

solve to get v as function of x: $$v(x) = \sqrt{\frac{c}{2m}(A^{4}- x^{4})}$$

then integrating of interval x = 0 to x = A:

$$T = 4 \int\frac{dx}{v(x)}$$

$$= 8A^{2}\sqrt{\frac{2m}{c}}$$

but it's wrong it just so wrong… please help
thanks!

 

[mighty2000]

Thank you for your post. I will do my best to help you with your problem. Here is my attempt at a solution:

a) To find the potential energy function, we can use the equation U = \int F dx. Plugging in the given force equation, we get:

U = \int -cx^3 dx = -\frac{1}{4}cx^4

Since we are taking U = 0 at x = 0, we can rewrite this as:

U = -\frac{1}{4}cx^4 + C

where C is a constant of integration. Since we are only interested in the potential energy function, we can ignore the constant and the final potential energy function is:

U = -\frac{1}{4}cx^4

b) To find the time for one-quarter of a period, we can use the equation T = \int\frac{dx}{v(x)}. Plugging in the given velocity equation, we get:

T = 4 \int_0^A \frac{dx}{\sqrt{\frac{c}{2m}(A^4 - x^4)}}

To solve this integral, we can use the substitution u = x^2, which gives us:

T = 4 \int_0^{A^2} \frac{du}{\sqrt{\frac{c}{2m}(A^4 - u^2)}}

This integral can be solved using a trigonometric substitution. Let u = Asin\theta, which gives us du = Acos\theta d\theta. Substituting this into the integral, we get:

T = 4 \int_0^{\frac{\pi}{2}} \frac{Acos\theta d\theta}{\sqrt{\frac{c}{2m}(A^4 - A^2sin^2\theta)}}

T = 4 \int_0^{\frac{\pi}{2}} \frac{Acos\theta d\theta}{\sqrt{\frac{c}{2m}A^2cos^2\theta}}

T = 4 \int_0^{\frac{\pi}{2}} \frac{Acos\theta d\theta}{\sqrt{\frac{c}{2m}A^2cos^2\theta}}

T = 4 \int_0^{\frac{\