Q133 / 180 · KCET 2014 · Physics
A train is approaching towards a platform with a speed of \( 10 \mathrm{~ms}^{-1} \) while blowing a whistle of
frequency \( 340 \mathrm{~Hz} \). What is the frequency of whistle heard by a stationary observer on the
platform ? Given speed of sound \( =340 \mathrm{~ms}^{-1} \).
frequency \( 340 \mathrm{~Hz} \). What is the frequency of whistle heard by a stationary observer on the
platform ? Given speed of sound \( =340 \mathrm{~ms}^{-1} \).
- A\( 330 \mathrm{~Hz} \)
- Correct answerB\( 350 \mathrm{~Hz} \)
- C\( 340 \mathrm{~Hz} \)
- D\( 360 \mathrm{~Hz} \)
Solution
According to Doppler effect there is a change in frequency or wavelength of a wave in relation to observer who is moving
relative to the wave source, which is given as \( f^{\prime}=\left(\frac{v}{v-v_{s}}\right) f \)
where \( \mathrm{f}^{\prime} \) is the apparent frequency; \( \mathrm{f} \) is frequency of source; \( v_{S} \) is velocity of source; \( v \) is velocity of sound
Given
\( f=340 H z ; v=340 \mathrm{~ms}^{-1} ; v_{s}=10 \mathrm{~ms}^{-1} \)
Therefore,
\( f^{\prime}=\left(\frac{340}{340-10}\right) \times 340=350.3030 H z \sim 350 H z \)
Thus, frequency of whistle heard by a stationary observer on the platform is \( 350 \mathrm{~Hz} \)
relative to the wave source, which is given as \( f^{\prime}=\left(\frac{v}{v-v_{s}}\right) f \)
where \( \mathrm{f}^{\prime} \) is the apparent frequency; \( \mathrm{f} \) is frequency of source; \( v_{S} \) is velocity of source; \( v \) is velocity of sound
Given
\( f=340 H z ; v=340 \mathrm{~ms}^{-1} ; v_{s}=10 \mathrm{~ms}^{-1} \)
Therefore,
\( f^{\prime}=\left(\frac{340}{340-10}\right) \times 340=350.3030 H z \sim 350 H z \)
Thus, frequency of whistle heard by a stationary observer on the platform is \( 350 \mathrm{~Hz} \)
Hence, the correct answer is option: 'B'.