# Amplitude of Oscillation Without Solving Differential Equation

I am currently working on problem in my own research. There seems to be a weak analogy between my problem and motion on a spring. Therefore, I am exploring this question in regards to a mass oscillating on a spring in hopes to gain further insight into my own system in question.

Here is the idea: We can write out the differential equation of motion for a mass on a spring

$m\ddot x=-kx$

Even though we can find an analytic solution to this equation, let's assume that we can only solve this equation numerically.

Let's say we start the mass at rest at some non-zero initial position. We know that the amplitude of the oscillation will be equal to the magnitude of the initial position (for example, if we start at x = 5 m, then the amplitude will be 5 m).

My question is this: Is there a way to determine that this is true of the amplitude without actually solving the differential equation. In other words, can we use the equation (and maybe other things we know about the system, like how at the maximum position the velocity is 0 and the acceleration will be at a maximum) to determine what will be our maximum position without actually solving the differential equation.

I know this is kind of vague, so if more information is needed please let me know.

• Perhaps to make this less vague you could give more details about the problem you are actually trying to solve. If you can solve the differential equation numerically then you can extract the quantities of interest from the numerical simulation. A general differential equation won't necessarily have a well defined "amplitude of oscillation". – jgerber May 16 '18 at 16:09

The answer is yes, because this equation of motion conserves energy. At any time, $E = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}kx^2$ is constant because $$\frac{dE}{dt} = \dot{x}\left(m\ddot{x}+kx\right) = 0$$
This means if we know the initial conditions $x(t=0)$ and $\dot{x}(t=0)$ we know the energy, and we can use that to find the position and velocity at later times. In particular, as $\dot{x}^2$ decreases $x^2$ must increase, and $\dot{x}^2$ can't be smaller than $0$ so $|x|$ is maximum when $$\frac{1}{2}kx^2 = E$$ or $$x_{max} = \sqrt{\frac{2E}{k}}$$
• Yes this makes sense thank you. What about if we included a damping term, say the typical $-b\dot x$. We cannot say the energy is constant, but it seems like we could make a similar argument knowing that at the maximum position we are not losing any energy to the damping. – Aaron Stevens May 16 '18 at 16:53
• @AaronStevens: If there is a damping term, then you can make the argument that $dE/dt = \dot{x} ( - b \dot{x})\leq 0$ at all times. The basic argument then proceeds with inequalities instead of equalities. – Michael Seifert May 16 '18 at 16:55
• In the special case that the damping force is constant you could use $1/2 k A^2 = E_0 - \int_{x_0}^A F_f dx = E_0 - F(A-x_0)$ where $F_f$ is your damping force. But this does not work for damping force like $F_f = -b\dot{x}$ because then the integral depends on the exact motion of the particle. So in general you have to start actually solving the equation of motion to get exact limits. – Luke Pritchett May 16 '18 at 17:46