1. Initial resonance condition

Let the original frequency of the stationary tuning fork be $f$.

Since the air column is in resonance with the tuning fork, the pipe length $L$ satisfies $L \propto \lambda$ where the wavelength in air is $\lambda = \frac{v}{f}$ with $v=320\,\text{m s}^{-1}$.

2. Effect of moving source

The tuning fork is moved parallel to the open end of the pipe. This means it is moving towards/away from the pipe along the direction of sound propagation at that end, so Doppler effect changes the frequency heard by the air column.

For a source moving with speed $u_s=2\,\text{m s}^{-1}$, the apparent frequency at the stationary pipe is

• when source moves towards pipe: $f' = f\frac{v}{v-u_s}$
• when source moves away from pipe: $f' = f\frac{v}{v+u_s}$

To get the smallest percentage change in pipe length, we choose the smaller fractional change, i.e. the case of source moving away, because then the frequency change is smaller in magnitude for the needed shortening/lengthening comparison.

3. New resonant length

For the same mode of resonance, pipe length is inversely proportional to frequency: $L' \propto \frac{1}{f'}$ Hence $\frac{L'}{L} = \frac{f}{f'}$

For the source moving away, $\frac{L'}{L} = \frac{f}{f\,v/(v+u_s)} = \frac{v+u_s}{v}$ $\frac{L'}{L} = \frac{320+2}{320} = \frac{322}{320} = 1.00625$

Therefore, $\frac{\Delta L}{L} = 1.00625-1 = 0.00625$

Percentage change: $0.00625\times 100 = 0.625\%$

4. Smallest percentage change

Thus the smallest required percentage change in the pipe length is $\boxed{0.625\%}$

This lies in the range $0.62$ to $0.63$.