In a resonance tube experiment, a tube of fixed length is closed at one end and several tunning forks of increasing frequency used to obtain resonance at the open end. If the tunning fork with the lowest frequency which gave resonance had a frequency f1 and the next tunning fork to give resonance had a frequency f2, find the ratio f2:f1

For a resonance tube closed at one end and open at the other, resonance occurs when the length of the tube $L$ is an odd multiple of a quarter wavelength. The resonant frequencies $f_n$ are given by:

$$f_n = \frac{nv}{4L}$$

where $n = 1, 3, 5, \dots$ (odd integers) and $v$ is the speed of sound in air.

The lowest frequency corresponds to $n=1$.

$$f_1 = \frac{1 \cdot v}{4L} = \frac{v}{4L}$$

The next resonant frequency corresponds to the next odd integer, which is $n=3$.

$$f_2 = \frac{3 \cdot v}{4L}$$

To find the ratio $\frac{f_2}{f_1}$, we divide the expression for $f_2$ by the expression for $f_1$:

$$\frac{f_2}{f_1} = \frac{\frac{3v}{4L}}{\frac{v}{4L}}$$ $$\frac{f_2}{f_1} = \frac{3v}{4L} \times \frac{4L}{v}$$ $$\frac{f_2}{f_1} = 3$$

So, the ratio $f_2:f_1$ is $3:1$.

The ratio $f_2:f_1$ is $\boxed{\text{3:1}}$.