Let's solve each part step by step, explaining the reasoning at each stage. --- ## (a) **Expression for \( L_x \) as a function of \( N \), \( \overline{u} \), and \( H \)** ### Step 1: **Gravity Wave Dispersion Relation** For internal gravity waves, the dispersion relation is: \[ \omega = \overline{u} k_x = \frac{N k_x}{\sqrt{k_x^2 + k_z^2}} \] Stationary waves mean the observed frequency is zero in the ground frame, so the intrinsic frequency is: \[ \omega - \overline{u} k_x = \implies \omega = \overline{u} k_x \] But for stationary waves, the phase speed must match the background flow: \[ c = \frac{\omega}{k_x} = \overline{u} \] ### Step 2: **Relating Vertical and Horizontal Wavelengths** Given \( L_z = H \), so \( k_z = \frac{2\pi}{H} \). From the dispersion relation for stationary waves: \[ \overline{u} k_x = \frac{N k_x}{\sqrt{k_x^2 + k_z^2}} \] Solving for \( k_x \): \[ \overline{u} = \frac{N}{\sqrt{1 + \left(\frac{k_z}{k_x}\right)^2}} \] \[ \overline{u}^2 = \frac{N^2}{1 + \left(\frac{k_z}{k_x}\right)^2} \] \[ 1 + \left(\frac{k_z}{k_x}\right)^2 = \frac{N^2}{\overline{u}^2} \] \[ \left(\frac{k_z}{k_x}\right)^2 = \frac{N^2}{\overline{u}^2} - 1 \] \[ k_x = \frac{k_z}{\sqrt{\frac{N^2}{\overline{u}^2} - 1}} \] The horizontal wavelength: \[ L_x = \frac{2\pi}{k_x} = \frac{2\pi}{k_z} \sqrt{\frac{N^2}{\overline{u}^2} - 1} \] But \( k_z = \frac{2\pi}{H} \), so: \[ L_x = H \sqrt{\frac{N^2}{\overline{u}^2} - 1} \] **Explanation**: We used the stationary gravity wave dispersion relation and the given vertical wavelength to get \( L_x \) in terms of \( N \), \( \overline{u} \), and \( H \). --- ## (b) **Minimum Height \( H \) for Stationary Gravity Waves** ### Step 1: **Condition for Real \( L_x \)** From part (a), the term under the square root must be positive: \[ \frac{N^2}{\overline{u}^2} - 1 > \implies \frac{N^2}{\overline{u}^2} > 1 \implies N > \overline{u} \] But let's solve for the minimum \( H \): From (a): \[ L_x = H \sqrt{\frac{N^2}{\overline{u}^2} - 1} \] For real \( L_x \), need \(\frac{N^2}{\overline{u}^2} - 1 > \). ### Step 2: **Solving for Minimum \( H \)** Alternatively, for a fixed \( L_x \), the minimum occurs as \( \frac{N^2}{\overline{u}^2} \to 1 \) from above. Let's solve for \( H \): Set denominator to : \[ \frac{N^2}{\overline{u}^2} - 1 = \implies H_\text{min} = \frac{2\pi \overline{u}}{N} \] But since \( k_z = \frac{2\pi}{H} \), the maximum \( k_z \) (minimum \( H \)) for stationary waves is when \( k_x \to \infty \), but for a finite \( L_x \), so: \[ H_\text{min} = \frac{2\pi \overline{u}}{N} \] **Explanation**: The minimum mountain height occurs where the stationary solution first becomes possible, i.e., when \( N = \overline{u} \). Solving for \( H \) gives the threshold value. --- ## (c) **Physical Explanation for the Minimum Height** If \( H \) is below the minimum value found in (b): - The vertical wavelength is too short for gravity waves to propagate as stationary solutions. - The frequency associated with such a short vertical wavelength exceeds the Brunt–Väisälä frequency \( N \), so buoyancy can't restore the displaced air parcels fast enough to support a wave. - Instead, any disturbance will be quickly mixed or dissipated rather than forming a propagating stationary wave. **Explanation**: For \( H \) below the threshold, the background stability is too weak (or the flow too fast) to sustain stationary gravity waves with that short vertical wavelength. --- ## **Final Answers** ### (a) \[ \boxed{ L_x = H \sqrt{\frac{N^2}{\overline{u}^2} - 1} } \] ### (b) \[ \boxed{ H_\text{min} = \frac{2\pi \overline{u}}{N} } \] ### (c) **For \( H < H_\text{min} \), stationary gravity waves cannot exist because the vertical wavelength is too short for buoyancy to act as a restoring force. The atmosphere cannot support waves with frequencies higher than the Brunt–Väisälä frequency \( N \), so the disturbance is not wavelike and is instead rapidly dissipated or mixed.**