C5-3. Equations for the Doppler Effect of SoundHL Only

1. Moving Source and Observer

When a source of sound waves moves relative to a stationary observer, the observed frequency can be calculated using the Doppler shift equation for a moving source. The wave velocity for sound waves is generally given as $340\text{ m s}^{-1}$.

Equation for a Moving Source:

$$f' = f \left( \frac{v}{v \pm u_s} \right)$$
  • The $\pm$ depends on whether the source is moving towards or away from the observer.
  • If the source is moving towards the observer, the denominator is $v - u_s$.
  • If the source is moving away from the observer, the denominator is $v + u_s$.

Equation for a Moving Observer:

When a source of sound waves remains stationary, but the observer is moving relative to the source, the observed frequency can be calculated using the Doppler shift equation for a moving observer:

$$f' = f \left( \frac{v \pm u_o}{v} \right)$$
  • The $\pm$ depends on whether the observer is moving towards or away from the source.
  • If the observer is moving towards the source, the numerator is $v + u_o$.
  • If the observer is moving away from the source, the numerator is $v - u_o$.

Equations in terms of Wavelength:
These equations can also be written in terms of wavelength. For example, the equation for a moving source is shown below:

$$\lambda' = \lambda \left( 1 \mp \frac{u_s}{v} \right)$$
  • If the source is moving towards, the term in the brackets is $1 - \dfrac{u_s}{v}$.
  • If the source is moving away, the term in the brackets is $1 + \dfrac{u_s}{v}$.
EXAMINER TIP

Pay careful attention as to whether you need to use the $+$ or $-$ sign in the relevant equation! If it helps, explicitly label the 'observer' and 'source' on your exam paper to make sure you use the correct $u_o$ or $u_s$.

Example 1: The Receding Police Siren

Problem: A police car siren emits a sound wave with a frequency of $450\text{ Hz}$. The car is travelling away from an observer at a speed of $45\text{ m s}^{-1}$. The speed of sound is $340\text{ m s}^{-1}$. What frequency of sound does the observer hear?
A. 519 Hz
B. 483 Hz
C. 397 Hz
D. 358 Hz

Solution:

The observer is stationary, and the source (police car) is moving away. Therefore, we use the moving source equation and add the velocities in the denominator to reduce the frequency:
$$ f' = f \left( \frac{v}{v + u_s} \right) = 450 \times \left( \frac{340}{340 + 45} \right) \approx 397.4\text{ Hz} $$ The correct answer is C. 397 Hz.

Example 2: The Escaping Robber

Problem: A bank robbery has occurred and the alarm is sounding at a frequency of $3\text{ kHz}$. The robber jumps into a car which accelerates and reaches a constant speed. As he drives away at a constant speed, he hears the frequency of the alarm decrease to $2.85\text{ kHz}$. Determine the speed at which the robber must be driving away from the bank. (Speed of sound = $340\text{ m s}^{-1}$).

Solution:

Step 1: Identify Equation
The observer is moving away from a stationary source of sound, so the equation to use is:
$$f' = f \left( \frac{v - u_o}{v} \right)$$

Step 2: Rearrange for observer velocity ($u_o$)
$$ \frac{f'}{f} = \frac{v - u_o}{v} \implies \frac{v f'}{f} = v - u_o \implies u_o = v - \frac{v f'}{f} = v \left(1 - \frac{f'}{f}\right) $$

Step 3: Calculate
$$ u_o = 340 \times \left(1 - \frac{2.85}{3}\right) = \mathbf{17\text{ m s}^{-1}} $$ The robber must be driving away at a constant speed of $17\text{ m s}^{-1}$ based on the change in frequency heard.

Example 3: HL Approaching Engine

Problem: A stationary observer is waiting at a crosswalk when a fire engine approaches them at a speed of $35\text{ m s}^{-1}$. The engine's siren has a true frequency of $800\text{ Hz}$. Calculate the frequency of the siren exactly as it is heard by the observer. (Assume the speed of sound is $340\text{ m s}^{-1}$).

Solution:

The observer is stationary, and the source is moving towards them. According to the Doppler principles, moving towards an observer compresses the wavefronts, increasing the frequency. We select the moving source equation and subtract the source velocity in the denominator to increase the overall value of the fraction:
$$f' = f \left( \frac{v}{v - u_s} \right)$$
Substitute the values:
$$ f' = 800 \times \left( \frac{340}{340 - 35} \right) = \mathbf{891.8\text{ Hz}} $$

2. Combined Motion and Double Doppler Effect (Reflections)

Combined Moving Source and Observer:

If both the source and the observer are moving simultaneously, the two effects multiply. We combine the equations into a single comprehensive master equation:

$$f' = f \left( \frac{v \pm u_o}{v \mp u_s} \right)$$
  • Evaluate the numerator (observer) and denominator (source) completely independently.
  • If a motion tends to bring them closer together, it must increase $f'$. Adjust the sign to make the fraction larger (i.e., $+u_o$ in numerator or $-u_s$ in denominator).
  • If a motion tends to pull them apart, it must decrease $f'$. Adjust the sign to make the fraction smaller (i.e., $-u_o$ in numerator or $+u_s$ in denominator).

Double Doppler Effect (Radar and Ultrasound):

When a stationary transmitter sends a wave (radio or ultrasound) at a moving target, the wave hits the target and bounces back. This causes a double Doppler shift:
1. The target acts as a moving observer receiving the wave.
2. The target then acts as a moving source reflecting the wave back.

For electromagnetic waves (like radar) where the target's speed $v_{\text{target}} \ll c$, the combined two-way frequency shift simplifies neatly to:

$$\Delta f \approx \frac{2v_{\text{target}}}{c} f$$

(Note: The factor of 2 accounts for the wave being shifted twice—once on the journey there, and once on the journey back).

Example 4: Radar Reflection (Double Doppler Shift)

Problem: A police stationary radar gun emits a continuous radio wave pulse with a frequency of $10.0\text{ GHz}$ ($10.0 \times 10^9\text{ Hz}$). A car is driving directly towards the radar gun at a speed of $33.0\text{ m s}^{-1}$. The wave bounces off the car and returns to the radar gun. Calculate the frequency shift ($\Delta f$) detected by the police radar. ($c = 3.00 \times 10^8\text{ m s}^{-1}$).

Solution:

Because the car is moving towards the radar, the wave undergoes a double blue-shift (once as the car intercepts the wave acting as an observer, and a second time as the car reflects the wave acting as a moving source).

For an electromagnetic wave reflecting off a non-relativistic target ($v \ll c$), the double shift is calculated using:
$$\Delta f \approx \frac{2v_{\text{target}}}{c} f$$
Substitute the known values:
$$\Delta f = \frac{2 \times 33.0}{3.00 \times 10^8} \times (10.0 \times 10^9) = 22.0 \times 100 = \mathbf{2200\text{ Hz} \text{ (or } 2.2\text{ kHz})} $$ The police radar detects a beat frequency (shift) of $2200\text{ Hz}$ and uses it to calculate the driver's speed.

Example 5: Combined Motion (Both Moving)

Problem: An ambulance is driving down a straight highway at $30\text{ m s}^{-1}$ with its siren emitting a frequency of $600\text{ Hz}$. A civilian car is driving ahead of the ambulance in the exact same direction at $20\text{ m s}^{-1}$. Assuming the speed of sound is $340\text{ m s}^{-1}$, what frequency does the driver of the civilian car hear?

Solution:

Step 1: Determine the signs for the Combined Equation
The combined equation is: $f' = f \left( \dfrac{v \pm u_o}{v \mp u_s} \right)$
$\bullet$ Observer (Car): The car is moving away from the ambulance. Moving away decreases frequency, so we make the numerator smaller and use $-u_o$.
$\bullet$ Source (Ambulance): The ambulance is moving towards the car. Moving towards increases frequency, so we make the denominator smaller and use $-u_s$.

Step 2: Setup the specific equation
$$f' = f \left( \frac{v - u_o}{v - u_s} \right)$$

Step 3: Calculate
$$ f' = 600 \times \left( \frac{340 - 20}{340 - 30} \right) = \mathbf{619.4\text{ Hz}} $$