A train is moving on a straight track with speed 20 m s$$^{-1}$$. It is blowing its whistle at the frequency of 1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound = 320 m s$$^{-1}$$) close to:

Let us denote the speed of sound in air by $$v$$, the speed of the train (which is the source of sound) by $$v_s$$, the actual frequency of the whistle by $$f$$, and the frequencies heard by the observer when the train approaches and when it recedes by $$f_1$$ and $$f_2$$ respectively.

Given data:

$$v = 320 \ {\rm m\,s^{-1}}, \qquad v_s = 20 \ {\rm m\,s^{-1}}, \qquad f = 1000 \ {\rm Hz}.$$

First we recall the Doppler effect formula for a stationary observer and a moving source. When the source approaches the observer, the apparent frequency is

$$f_1 = f \,\frac{v}{\,v - v_s\,}.$$

When the source moves away (recedes) from the observer, the apparent frequency is

$$f_2 = f \,\frac{v}{\,v + v_s\,}.$$

We now substitute the numerical values.

For the approaching train:

$$f_1 = 1000 \times \frac{320}{320 - 20} = 1000 \times \frac{320}{300} = 1000 \times 1.066666\ldots = 1066.666\ldots \ {\rm Hz}.$$

For the receding train:

$$f_2 = 1000 \times \frac{320}{320 + 20} = 1000 \times \frac{320}{340} = 1000 \times 0.941176\ldots = 941.176\ldots \ {\rm Hz}.$$

The change in the heard frequency as the train passes is

$$\Delta f = f_1 - f_2 = 1066.666\ldots - 941.176\ldots = 125.490\ldots \ {\rm Hz}.$$

To find the percentage change, we compare this change with the original (real) frequency $$f = 1000 \,{\rm Hz}.$$ Thus

$$\text{Percentage change} = \frac{\Delta f}{f}\times 100\% = \frac{125.490\ldots}{1000}\times 100\% = 12.549\ldots\%.$$

This is approximately $$12\%.$$

Hence, the correct answer is Option C.