An organ pipe 40 cm long is open at both ends. The speed of sound in air is 360 m s$$^{-1}$$. The frequency of the second harmonic is ________ Hz.

We need to find the frequency of the second harmonic in an open pipe.

Formula for harmonics in an open pipe. For a pipe open at both ends, the frequency of the $$n$$th harmonic is:

$$f_n = \frac{nv}{2L}$$

where $$v$$ is the speed of sound and $$L$$ is the length of the pipe.

Given values: $$L = 40$$ cm $$= 0.4$$ m, $$v = 360$$ m/s, $$n = 2$$ (second harmonic)

Calculate: $$f_2 = \frac{2 \times 360}{2 \times 0.4} = \frac{720}{0.8} = 900 \text{ Hz}$$

The correct answer is 900 Hz.