Question

A train moving at 20 m/s approaches a stationary observer. The frequency of the whistle emitted by the train is 640 Hz. If the velocity of sound is 340 m/s, the apparent frequency heard by the observer is:

• A. 680 Hz
• B. 600 Hz
• C. 720 Hz
• D. 640 Hz

Hint:

Identify which object is moving, the source or the observer, and in which direction, since this decides whether you add or subtract the object speed in the Doppler formula. Keep the speed of sound and every object speed in the same units before substituting.

Answer 1

The Correct Option is A

Step 1: Concept
This uses the Doppler Effect. When a source approaches a stationary observer, the apparent frequency is higher than the emitted frequency.

Step 2: Meaning
The Doppler formula for a moving source and stationary observer: $f_{app} = f_0\!\left(\dfrac{v}{v - v_s}\right)$, where $v$ is the speed of sound and $v_s$ is the source speed.

Step 3: Analysis
With $f_0 = 640$ Hz, $v = 340$ m/s, $v_s = 20$ m/s: \[f_{app} = 640 \times \frac{340}{340-20} = 640 \times \frac{340}{320} = 640 \times \frac{17}{16} = 680\ \text{Hz}.\]

Step 4: Conclusion
The apparent frequency heard by the observer is 680 Hz. Final Answer: (A)

Answer 2

Concept:

• Instead of applying the Doppler formula directly, the apparent frequency can be derived by first finding how the wavelength of sound changes because the source is moving toward the observer.
• In one time period of the wave, the source moves closer to the observer, compressing the wave crests still ahead of it into a shorter distance. This shorter distance becomes the new, compressed wavelength.

Step 1: Find the original time period and wavelength of the sound.
$T_0 = \dfrac{1}{f_0} = \dfrac{1}{640}\ \text{s}$
$\lambda_0 = \dfrac{v}{f_0} = \dfrac{340}{640}\ \text{m}$

Step 2: Find the compressed wavelength ahead of the moving source.
In time $T_0$, the wavefront emitted at the start of the period has moved a distance $vT_0$ ahead, while the source itself has moved a distance $v_sT_0$ toward the observer. The new wavelength is the gap between these two positions.
$\lambda_1 = vT_0 - v_sT_0 = (v-v_s)T_0 = \dfrac{v-v_s}{f_0}$
$\lambda_1 = \dfrac{340-20}{640} = \dfrac{320}{640}\ \text{m}$

Step 3: Convert the compressed wavelength back into a frequency.
$f_{app} = \dfrac{v}{\lambda_1} = \dfrac{v}{(v-v_s)/f_0} = f_0\dfrac{v}{v-v_s}$
$f_{app} = 640\times\dfrac{340}{320} = 680\ \text{Hz}$

Final Answer: 680 Hz