Q157 / 200 · MHT CET - Medical 2026 (23 April Shift 1) · Physics
A train moves towards a stationary observer with speed $20$ m/s. The train sounds a whistle and its frequency heard by the observer is $f_1$. When the train moves away from the platform at a speed of $20$ m/s, the frequency heard is $f_2$. If the speed of sound is $340$ m/s, then $\dfrac{f_2}{f_1}$ is
- A$\dfrac{7}{8}$
- B$\dfrac{8}{7}$
- C$\dfrac{9}{8}$
- Correct answerD$\dfrac{8}{9}$
Solution
Using Doppler's effect, the apparent frequency when the source moves towards a stationary observer is given by:
$f_1 = f_0 \left( \dfrac{v}{v - v_s} \right)$
Substituting $v = 340$ m/s and $v_s = 20$ m/s:
$f_1 = f_0 \left( \dfrac{340}{340 - 20} \right) = f_0 \left( \dfrac{340}{320} \right)$
When the source moves away from the stationary observer, the apparent frequency is given by:
$f_2 = f_0 \left( \dfrac{v}{v + v_s} \right)$
Substituting the values:
$f_2 = f_0 \left( \dfrac{340}{340 + 20} \right) = f_0 \left( \dfrac{340}{360} \right)$
Taking the ratio $\dfrac{f_2}{f_1}$:
$\dfrac{f_2}{f_1} = \dfrac{f_0 \left( \dfrac{340}{360} \right)}{f_0 \left( \dfrac{340}{320} \right)} = \dfrac{320}{360} = \dfrac{8}{9}$
Answer: $\dfrac{8}{9}$
$f_1 = f_0 \left( \dfrac{v}{v - v_s} \right)$
Substituting $v = 340$ m/s and $v_s = 20$ m/s:
$f_1 = f_0 \left( \dfrac{340}{340 - 20} \right) = f_0 \left( \dfrac{340}{320} \right)$
When the source moves away from the stationary observer, the apparent frequency is given by:
$f_2 = f_0 \left( \dfrac{v}{v + v_s} \right)$
Substituting the values:
$f_2 = f_0 \left( \dfrac{340}{340 + 20} \right) = f_0 \left( \dfrac{340}{360} \right)$
Taking the ratio $\dfrac{f_2}{f_1}$:
$\dfrac{f_2}{f_1} = \dfrac{f_0 \left( \dfrac{340}{360} \right)}{f_0 \left( \dfrac{340}{320} \right)} = \dfrac{320}{360} = \dfrac{8}{9}$
Answer: $\dfrac{8}{9}$
Hence, the correct answer is option: 'D'.