Here, frequency of source of sound, $\upsilon = 400$ Hz; $v = 340\ \mathrm{m\,s^{-1}}$ = speed of sound Speed of source $= v_s = 10\ \mathrm{m\,s^{-1}}$ (i) (a) When the train approaches the platform, the apparent frequency as heard by the observer on the platformwill be $\upsilon' = \frac{v}{v-v_{s}} \upsilon = \frac{340}{340-10} \times 400$ $= \frac{340}{330} \times 400 = 412.1\ \mathrm{Hz}$ (b) When the train recedes from the platform, the apparent frequency as heard by the observer is given by according to the formula :$\upsilon' = \frac{v}{v+v_{s}} \upsilon = \frac{340}{340+10} \times 400$$= \frac{340}{350} \times 400 = 388.6\ \mathrm{Hz} = 389\ \mathrm{Hz}$(ii) The speed of sound in each case remains same i.e. $340\ \mathrm{m\,s^{-1}}$.