A train is moving on a straight track with speed 20m s ^ - 1 . It...

Before train passes

${f}^{′} = \left(\dfrac{v}{v - {v}_{s}}\right) \times {f}_{0}$

$= \left(\dfrac{320}{320 - 20}\right) \times {f}_{0}$

$= \left(\dfrac{320}{300}\right) \times {f}_{0}$

$= \dfrac{16{f}_{0}}{15}$

After train passes

${f}^{′′} = \left(\dfrac{v}{v + {v}_{s}}\right) \times {f}_{0}$

${f}^{′′} = \left(\dfrac{320}{320 + 20}\right){f}_{0}$

$= \left(\dfrac{320}{340}\right){f}_{0}$

$= \left(\dfrac{16}{17}\right){f}_{0}$

Change in frequency

$= \left(\dfrac{16}{15} - \dfrac{16}{17}\right){f}_{0}$

$∴$ Percentage change in frequency

$= \dfrac{\left(\dfrac{16}{15} - \dfrac{16}{17}\right){f}_{0}}{{f}_{0}} \times 100 \approx 12$ nearly