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(A) Using Doppler's effect formula:$1$) Frequency when the train approaches: $f_{	ext{app}} = f_0 	imes \frac{v}{v - v_s}$$2$) Frequency when the tr

Vedclass · Accepted Answer

$18.33$

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(A) Using Doppler's effect formula:$1$) Frequency when the train approaches: $f_{	ext{app}} = f_0 	imes \frac{v}{v - v_s}$$2$) Frequency when the train recedes: $f_{	ext{rec}} = f_0 	imes \frac{v}{v + v_s}$Given: $v = 330 	ext{ m/s}$, $v_s = 108 	ext{ km/h} = 108 	imes \frac{5}{18} = 30 	ext{ m/s}$.The difference in frequency is $\Delta f = f_{	ext{app}} - f_{	ext{rec}} = f_0 \left( \frac{v}{v - v_s} - \frac{v}{v + v_s} ight)$.$\Delta f = f_0 \left( \frac{v(v + v_s) - v(v - v_s)}{v^2 - v_s^2} ight) = f_0 \left( \frac{2 v v_s}{v^2 - v_s^2} ight)$.Percentage change = $\frac{\Delta f}{f_0} 	imes 100 = \left( \frac{2 v v_s}{v^2 - v_s^2} ight) 	imes 100$.Substituting the values: $\frac{2 	imes 330 	imes 30}{330^2 - 30^2} 	imes 100 = \frac{19800}{108900 - 900} 	imes 100 = \frac{19800}{108000} 	imes 100 = \frac{11}{60} 	imes 100 \approx 18.33 \%$.

$A$ boy standing on a platform observes the frequency of a train horn as it passes by. The change in the frequency noticed as the train approaches and recedes him with a velocity of $108 \text{ km/h}$ is (speed of sound in air $= 330 \text{ m/s}$) (in $\%$)

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