What is "equilibrium position" in simple harmonic motion?

Question

In simple harmonic motion (SHM) is equilibrium position equal to the extreme position (i.e. where the external force and restoring force are equal), or where all kinetic energy of the body is converted into the potential energy?

Answer 1

TL;DR In the context of first Newton's law, the equilibrium position is a position where vector sum of all forces equals zero. But beware, whether or not the object will actually rest at that position depends on nature of system, initial conditions and external forces.

is equilibrium position equal to the extreme position (i.e. where the external force and restoring force are equal)

I do not know what do you mean by "extreme position", but equilibrium position does not need to be equal to the spring maximum elongation. To explain what I mean, I discuss below two cases - a spring-mass system in horizontal and vertical configuration. The former does not have any external force, while the latter naturally has the gravitational force acting on the object.

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Spring-mass system in horizontal configuration

The differential equation that describes motion of the mass is

$$\ddot{x}(t) + \omega_0^2 x(t) = u(t)$$

where $x$ denotes the spring elongation, $\omega_0 = \sqrt{k/m}$ is frequency of oscillations (aka the natural frequency), and $u$ is contribution from external force(s). When there are no external forces ($u = 0$), the rest position is at $x_\text{eq} = 0$ and the solution to the above differential equation is

$$x(t) = x_0 \cos(\omega_0 t) + \frac{\dot{x}_0}{\omega_0} \sin(\omega_0 t)$$

where $x_0 = x(0)$ and $\dot{x}_0 = \dot{x}(0)$ are initial mass position (spring elongation) and velocity. The above equation can also be written as

$$\boxed{x(t) = A \sin(\omega_0 t + \varphi)}$$

where amplitude $A$ and phase $\varphi$ are

$$A = \sqrt{x_0^2 + \frac{\dot{x}_0^2}{\omega_0^2}} \qquad \text{and} \qquad \varphi = \arctan (\frac{\omega_0 x_0}{\dot{x}_0})$$

If at least one of the initial conditions ($x_0$, $\dot{x}_0$) is not zero, the mass will permanently oscillate. This means that although the system rest position is at $x_\text{eq} = 0$, the mass will actually rest at that position only for zero initial conditions.

You can also analyze this from the work-energy perspective. Sum of kinetic energy $K$ and elastic potential energy $U_e$ equals mechanical energy

$$\boxed{E = K + U_e = \text{const.}}$$

which is constant in absence of external forces such as friction. When

• mass goes through position of zero elongation ($x = 0$), all mechanical energy is in the kinetic energy and mass has maximum velocity;

• mass reaches maximum spring elongation $x = \pm A$, all mechanical energy is in the elastic potential energy and mass has zero velocity;

How does mass start oscillating? There must have been some external force which gave the system initial mechanical energy (i.e., non-zero initial conditions). But this external force is not necessary to sustain oscillations, because mechanical energy is just continuously distributed between kinetic energy and elastic potential energy. In reality, there is always some (kinetic) friction force which dissipates mechanical energy and the mass will finally settle at $x = 0$.

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Spring-mass system in vertical configuration

The reasoning is similar to the spring-mass system in horizontal configuration, but the differential equation is slightly different:

$$\ddot{x}(t) + \omega_0^2 x(t) = -g + u(t)$$

where $x$ denotes the spring elongation, positive $x$ direction points upwards (away from Earth's center), and $u$ is contribution from external force(s) other than gravitational force. The system rest position is at $x_\text{eq} = -g/\omega_0^2$, but it does not mean mass will actually rest at this position.

If we introduce new variable $z(t) = x(t) + g/\omega_0^2$, the above differential equation becomes

$$\ddot{z}(t) + \omega_0^2 z(t) = u(t)$$

where $z$ denotes distance from the rest position, and the solution is

$$z(t) = A \sin(\omega_0 t + \varphi)$$

where amplitude $A$ and phase $\varphi$ are

$$A = \sqrt{z_0^2 + \frac{\dot{z}_0^2}{\omega_0^2}} \qquad \text{and} \qquad \varphi = \arctan (\frac{\omega_0 z_0}{\dot{z}_0})$$

Notice the similarity with solution to horizontal spring-mass system configuration. The only difference is that mass could rest at a position which does not equal position in which spring is neither stretched nor compressed. But the mass will rest only if initial conditions are zero!

Answer 2

The equilibrium refers to the point where the sum of all forces are balanced net force equals zero and potential energy has local minima or maxima. Now perhaps you are confused about the application of external force, then sure, when force is still present, then the body is in equilibrium, but when force is withdrawn, the body starts producing SHM. Now only restoring (internal) force acts.

At this instant, the position/point where restoring force is zero, that point is equilibrium point. $F_{\text{restoring}}=-kx$, So at $x=0$, $F_{\text{restoring}}=0$.

Now note also that potential energy will be minimum here: $U=1/2mA^2\omega ^2\sin^2(\omega t+\phi)$ For phase angle zero, $U=0$, so, minima exist.

Answer 3

the equation of motion in your case is:

$$m\,\ddot x+k\,x=f(t)$$

thus for the steady state $~\ddot x=0~$

$$f(t)=k,x(t)\quad \Rightarrow x_s=\frac{f(t)}{k}$$

where $~x_s~$ is the equilibrium state.

you don't obtain this result with the energy equation