A musician using an open flute of length 50 cm produces second harmonic sound waves. A person runs towards the musician from another end of a hall at a speed of 10 km h$$^{-1}$$. If the wave speed is 330 m s$$^{-1}$$, the frequency heard by the running person shall be close to:

We have an open flute, i.e. an organ pipe open at both ends, whose length is given as $$L = 50\ \text{cm} = 0.50\ \text{m}$$. An open pipe supports standing waves that have antinodes at both ends. For such a pipe, the allowed natural (harmonic) frequencies are obtained from the well-known relation

$$f_n = n\,\frac{v}{2L},\qquad n = 1,2,3,\ldots$$

where $$v$$ is the speed of sound in air and $$n$$ is the harmonic number. Here the musician is producing the second harmonic, so we put $$n = 2$$:

$$f_{\text{source}} = f_2 = 2\,\frac{v}{2L} = \frac{v}{L}.$$

Substituting the numerical values $$v = 330\ \text{m s}^{-1}$$ and $$L = 0.50\ \text{m}$$, we get

$$f_{\text{source}} = \frac{330\ \text{m s}^{-1}}{0.50\ \text{m}} = 660\ \text{Hz}.$$

This 660 Hz is the frequency emitted by the flute. Next, a listener (observer) is running directly toward the stationary source with a speed of $$10\ \text{km h}^{-1}$$. First we convert this speed into SI units:

$$10\ \text{km h}^{-1} = 10\times\frac{1000\ \text{m}}{3600\ \text{s}} = 2.777\ \text{m s}^{-1}\approx 2.78\ \text{m s}^{-1}.$$

For the Doppler effect when the observer moves toward a stationary source, the observed frequency $$f_{\text{obs}}$$ is given by the formula

$$f_{\text{obs}} = f_{\text{source}}\left(1 + \frac{v_{\text{observer}}}{v}\right),$$

because the wavelength remains the same while the observer meets the wavefronts more frequently. Substituting the known quantities:

$$f_{\text{obs}} = 660\ \text{Hz}\left(1 + \frac{2.78\ \text{m s}^{-1}}{330\ \text{m s}^{-1}}\right).$$

First compute the fractional term:

$$\frac{2.78}{330} \approx 0.008424.$$

Adding 1 gives

$$1 + 0.008424 = 1.008424.$$

Now multiply by the emitted frequency:

$$f_{\text{obs}} = 660\ \text{Hz}\times 1.008424 \approx 665.56\ \text{Hz}.$$

Rounding to the nearest whole number, the frequency heard by the running person is approximately $$666\ \text{Hz}$$, which matches option C.

Hence, the correct answer is Option C.