# Motion of a mass on an ideal spring [closed]

Consider the following figure :

Now my question : In the second case let $$x_1 \neq x _{\text{max}} > 0$$ In the third case It can be said that $$x_2 =-x_1$$ ? or must be $$x_2 =-x _{\text{max}}$$? and why ?

## closed as off-topic by Aaron Stevens, John Rennie, ja72, M. Enns, ZeroTheHeroNov 19 '18 at 3:55

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• Many answers are assuming oscillations when you have specified none. Can you please clarify your question? – Aaron Stevens Nov 17 '18 at 17:03

If $$x=x_1=-x_2$$ and $$x_1\neq x_{max}$$ then $$x_2\neq -x_{max}$$. I didnt understand what confuses you here.

The Question is a little bit confusing, let me explain, for your system the Equation of motion is,

$$\frac{d^2x}{dt^2}=-\frac{k}{m}x$$; and $$\frac{k}{m}=\omega^{2}$$

Here $$k$$ is the force constant of the spring and $$m$$ is the mass of the particle.

The solution is $$x=x_{max}sin(\omega t)$$; according to your notation(at initial time,$$t = 0; x = 0$$). Now different t will give you different displacement about the origin $$x=0$$. I think thinking in terms of the oscillatory solution will help you to understand the phenomena better.

Neither have to be true. The diagram is just showing the sign conventions being used. Nowhere in the diagram do they mention how the displacements relate to the "maximum" displacement (I'm not sure what maximum you refer to here), nor does it say how the two displacements relate to each other.

Or are you really just asking if we make the mass oscillate will it oscillate around $$x=0$$? I ask because the diagram has nothing about oscillations, and your question doesn't ask about oscillations either. Also the velocity of the mass is not specified in any of the diagrams

The system will have simple harmonic motion.

In mechanics and physics, simple harmonic motion is a special type of periodic motion or oscillation motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement.

https://en.wikipedia.org/wiki/Simple_harmonic_motion

As in this system $$F=-kx$$
This implies that this will have a simple harmonic motion. Where, k is spring constant.

Deformation of a simple harmonic oscillator is a form of potential energy given by:

$$PE =\frac12kx^2$$

Because a simple harmonic oscillator has no dissipative forces, the other important form of energy is kinetic energy.

Conservation of energy for these two forms at any displacement $$x$$ is:

\begin{align} KE + PE &= \rm constant\\ \frac12mv^2 + \frac12kx^2 &= \rm constant \end{align}

The maximum velocity will be at $$x=0$$ (equilibrium position).
The maximum potential energy will be at $$x_{\rm max}$$.

At any two points $$x_1$$ and $$x_2$$, where kinetic energy is equal, $$x_1=-x_2$$.