An observer is riding on a bicycle and moving towards a hill at $$18 \text{ km h}^{-1}$$. He hears a sound from a source at some distance behind him directly as well as after its reflection from the hill. If the original frequency of the sound as emitted by source is $$640 \text{ Hz}$$ and velocity of the sound in air is $$320 \text{ m s}^{-1}$$, the beat frequency between the two sounds heard by observer will be ______ Hz.

Given: Observer speed $$v_o = 18 \text{ km h}^{-1} = 5 \text{ m s}^{-1}$$ (moving towards the hill), source frequency $$f_0 = 640 \text{ Hz}$$, speed of sound $$v = 320 \text{ m s}^{-1}$$.

The observer hears two sounds: the direct sound from the source behind him, and the reflected sound from the hill ahead.

Direct sound (source behind, observer moving away from source):

$$f_1 = f_0 \left(\dfrac{v - v_o}{v}\right) = 640 \times \dfrac{320 - 5}{320} = 640 \times \dfrac{315}{320}$$

Reflected sound (hill acts as a stationary source ahead, observer moving towards it):

The hill reflects the original frequency $$f_0$$ (since the source is stationary). The observer moves toward the hill, so:

$$f_2 = f_0 \left(\dfrac{v + v_o}{v}\right) = 640 \times \dfrac{320 + 5}{320} = 640 \times \dfrac{325}{320}$$

Beat frequency:

$$f_{\text{beat}} = f_2 - f_1 = 640 \times \dfrac{325 - 315}{320} = 640 \times \dfrac{10}{320} = \dfrac{6400}{320} = 20 \text{ Hz}$$

Therefore, the beat frequency is $$\boxed{20}$$ Hz.