I have read in many places that below resonance the driving force is in phase with the harmonic oscillator. I have also read that the driving oscillator is in phase with the harmonic oscillator, however, I would not expect both to be true.

Back to basics:

Let the position of the 'wall' be given by

$$x_w(t) = X_{max}\cos\omega t$$

The force on the mass is then

$$F_m(t) = m\ddot x_m = -k[(x_m(t) - x_0) - x_w(t)]$$

where $x_0$ is the relaxed length of the spring. The differential equation is then

$$\ddot x_m + \frac{k}{m}x_m = \frac{k}{m}[x_0 + x_w(t)]$$

The transfer function, $\frac{X_m(\omega)}{X_w(\omega)}$, is then (for $\omega > 0$)

$$\frac{X_m(\omega)}{X_w(\omega)} = \frac{\omega^2_0}{\omega^2_0 - \omega^2},\qquad \omega_0 \equiv \sqrt{\frac{k}{m}}$$

It is easy to see that the wall and mass displacements are in phase for $\omega < \omega_0$ and in anti-phase for $\omega > \omega_0$.

If the 'driving force' is understood as the forcing function on the right hand side of the differential equation then, being proportional to $x_w(t)$, it too is in phase for $\omega < \omega_0$.

I have read in many places that below resonance the driving force is in phase with the harmonic oscillator. I have also read that the driving oscillator is in phase with the harmonic oscillator, however, I would not expect both to be true.

Back to basics:

Let the position of the 'wall' be given by

$$x_w(t) = X_{max}\cos\omega t$$

The force on the mass is then

$$F_m(t) = m\ddot x_m = -k[(x_m(t) - x_0) - x_w(t)]$$

where $x_0$ is the relaxed length of the spring. The differential equation is then

$$\ddot x_m + \frac{k}{m}x_m = \frac{k}{m}[x_0 + x_w(t)]$$

The transfer function, $\frac{X_m(\omega)}{X_w(\omega)}$, is then (for $\omega > 0$)

$$\frac{X_m(\omega)}{X_w(\omega)} = \frac{\omega^2_0}{\omega^2_0 - \omega^2},\qquad \omega_0 \equiv \sqrt{\frac{k}{m}}$$

It is easy to see that the wall and mass displacements are in phase for $\omega < \omega_0$ and in anti-phase for $\omega > \omega_0$.

$$x_m(t) = \frac{\omega^2_0}{\omega^2_0 - \omega^2}X_{max}\cos\omega t$$

If the 'driving force' is understood as the forcing function on the right hand side of the differential equation then, being proportional to $x_w(t)$, it too is in phase for $\omega < \omega_0$.

Assuming the (angular) frequency of the sinusoidal driving force, $F(t) = F_0\cos\omega t$ is below the resonance frequency $\omega_0$,

I have read in many places that below resonance the driving force is in phase with the harmonic oscillator. I have also read that the driving oscillator is in phase with the harmonic oscillator, however, I would not expect both to be true.

Back to basics:

Let the position of the 'wall' be given by

$$x_w(t) = X_{max}\cos\omega t$$

The force on the mass is then

$$F_m(t) = m\ddot x_m = -k[(x_m(t) - x_0) - x_w(t)]$$

where $x_0$ is the relaxed length of the spring. The differential equation is then

$$\ddot x_m + \frac{k}{m}x_m = \frac{k}{m}[x_0 + x_w(t)]$$

The transfer function, $\frac{X_m(\omega)}{X_w(\omega)}$, is then (for $\omega > 0$)

$$\frac{X_m(\omega)}{X_w(\omega)} = \frac{\omega^2_0}{\omega^2_0 - \omega^2},\qquad \omega_0 \equiv \sqrt{\frac{k}{m}}$$

It is easy to see that the wall and mass displacements are in phase for $\omega < \omega_0$ and in anti-phase for $\omega > \omega_0$.

If the 'driving force' is understood as the forcing function on the right hand side of the differential equation then, being proportional to $x_w(t)$, it too is in phase for $\omega < \omega_0$.

I have read in many places that below resonance the driving force is in phase with the harmonic oscillator. I have also read that the driving oscillator is in phase with the harmonic oscillator, however, I would not expect both to be true.

Back to basics:

Let the position of the 'wall' be given by

$$x_w(t) = X_{max}\cos\omega t$$

The force on the mass is then

$$F_m(t) = m\ddot x_m = -k[(x_m(t) - x_0) - x_w(t)]$$

where $x_0$ is the relaxed length of the spring. The differential equation is then

$$\ddot x_m + \frac{k}{m}x_m = \frac{k}{m}[x_0 + x_w(t)]$$

The transfer function, $\frac{X_m(\omega)}{X_w(\omega)}$, is then (for $\omega > 0$)

$$\frac{X_m(\omega)}{X_w(\omega)} = \frac{\omega^2_0}{\omega^2_0 - \omega^2},\qquad \omega_0 \equiv \sqrt{\frac{k}{m}}$$

It is easy to see that the wall and mass displacements are in phase for $\omega < \omega_0$ and in anti-phase for $\omega > \omega_0$.

If the 'driving force' is understood as the forcing function on the right hand side of the differential equation then, being proportional to $x_w(t)$, it too is in phase for $\omega < \omega_0$.