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Let's break down and solve this problem step by step.

### Given:
- The system is a spring-mass oscillator.
- Displacement \( s \) measures the mass's movement from the unstretched spring position.
- The acceleration is given by \( a = -4s - 2s^3 \) m/s².
- The initial velocity of the mass is \( v = 1 \) m/s at \( s = 0 \) (unstretched position).

We are tasked to find:
1. **(a)** How far the mass will move before coming to a stop.
2. **(b)** The velocity of the mass when it returns to the position \( s = 0 \).

### Part (a) - Maximum Displacement:

To find how far the mass moves before it stops, we'll use the work-energy principle.

The acceleration is a function of displacement \( s \), so we can express it as:
\[
a = v \frac{dv}{ds}
\]
Given \( a = -4s - 2s^3 \), equating the two:
\[
v \frac{dv}{ds} = -4s - 2s^3
\]
We can now separate variables:
\[
v dv = (-4s - 2s^3) ds
\]
Integrate both sides:
\[
\int v dv = \int (-4s - 2s^3) ds
\]
The left-hand side integrates to:
\[
\frac{v^2}{2}
\]
The right-hand side integrates to:
\[
-2s^2 - \frac{s^4}{2}
\]
Thus, the equation becomes:
\[
\frac{v^2}{2} = -2s^2 - \frac{s^4}{2} + C
\]
Now, using the initial conditions \( s = 0 \) and \( v = 1 \, \text{m/s} \), we can find the constant \( C \):
\[
\frac{(1)^2}{2} = -2(0)^2 - \frac{(0)^4}{2} + C \quad \Rightarrow \quad C = \frac{1}{2}
\]
So, the equation of motion becomes:
\[
\frac{v^2}{2} = -2s^2 - \frac{s^4}{2} + \frac{1}{2}
\]
To find the maximum displacement (when \( v = 0 \)), set \( v = 0 \) and solve for \( s \):
\[
0 = -2s^2 - \frac{s^4}{2} + \frac{1}{2}
\]
Multiply through by 2 to simplify:
\[
0 = -4s^2 - s^4 + 1
\]
Rearrange:
\[
s^4 + 4s^2 - 1 = 0
\]
Let \( u = s^2 \), so we get a quadratic equation:
\[
u^2 + 4u - 1 = 0
\]
Solve this using the quadratic formula:
\[
u = \frac{-4 \pm \sqrt{16 + 4}}{2} = \frac{-4 \pm \sqrt{20}}{2} = \frac{-4 \pm 2\sqrt{5}}{2}
\]
\[
u = -2 \pm \sqrt{5}
\]
Since \( u = s^2 \) must be non-negative, we take the positive root:
\[
u = -2 + \sqrt{5}
\]
Thus:
\[
s^2 = -2 + \sqrt{5}
\]
\[
s = \sqrt{-2 + \sqrt{5}} \quad (\text{maximum displacement})
\]

### Part (b) - Velocity at \( s = 0 \):

We return to the equation of motion when \( s = 0 \):
\[
\frac{v^2}{2} = -2(0)^2 - \frac{(0)^4}{2} + \frac{1}{2}
\]
This gives \( v = 1 \, \text{m/s} \), meaning the velocity is the same as the initial velocity.

Would you like further details or explanations on any step?

Here are 5 related questions to expand on this:
1. How would the solution change if the acceleration were linear instead of nonlinear?
2. Can we solve the motion equation using energy conservation principles directly?
3. How would damping or friction affect the maximum displacement and velocity?
4. What role does the cubic term in the acceleration equation play in the oscillatory behavior?
5. How can we numerically solve the fourth-degree polynomial for a more exact value?

**Tip:** When faced with nonlinear equations like these, numerical methods can provide precise answers when algebraic solutions are difficult to obtain.

A spring-mass oscillator consists of a mass and a spring connected as shown. The coordinate s measures the displacement of the mass relative to its position when the spring is unstretched. Suppose that the nonlinear spring subjects the mass to an acceleration a = -4s - 2s^3 m/s² and that you give the mass a velocity v = 1 m/s in the position s = 0. (a) How far will the mass move to the right before the springs bring it to a stop? (b) What will be the velocity of the mass when it has returned to the position s = 0?

Explore the motion of a spring-mass oscillator subjected to nonlinear acceleration a = -4s - 2s^3 m/s². The problem involves solving differential equations and using the work-energy principle to find maximum displacement and return velocity. Suitable for high school and college physics students.

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