A car sounding a horn producing a note of 500 Hz., approaches and then passes a stationary observer at a steady speed of `15ms^(-1)?`. Calculate the change in pitch of the note heard by the observer. (Velocity of sound `= 340 ms^(-1))`.

Answer

To solve the problem of the change in pitch of the note heard by the observer when a car approaches and then passes by, we can use the Doppler effect formula. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Frequency of the horn, \( f_0 = 500 \, \text{Hz} \)
- Speed of the car (source), \( v_s = 15 \, \text{m/s} \)
- Speed of sound, \( v = 340 \, \text{m/s} \)
- Speed of the observer, \( v_o = 0 \, \text{m/s} \) (since the observer is stationary)

### Step 2: Calculate the frequency heard by the observer when the source is approaching
Using the Doppler effect formula for a source approaching a stationary observer:
\[ f_1 = f_0 \frac{v + v_o}{v - v_s} \]
Substituting the known values:
\[ f_1 = 500 \frac{340 + 0}{340 - 15} = 500 \frac{340}{325} \]
Calculating \( f_1 \):
\[ f_1 = 500 \times 1.04615 \approx 523.08 \, \text{Hz} \]

### Step 3: Calculate the frequency heard by the observer when the source is receding
Using the Doppler effect formula for a source moving away from a stationary observer:
\[ f_2 = f_0 \frac{v + v_o}{v + v_s} \]
Substituting the known values:
\[ f_2 = 500 \frac{340 + 0}{340 + 15} = 500 \frac{340}{355} \]
Calculating \( f_2 \):
\[ f_2 = 500 \times 0.95775 \approx 478.87 \, \text{Hz} \]

### Step 4: Calculate the change in pitch
The change in pitch \( \Delta f \) is given by the difference between the two frequencies:
\[ \Delta f = f_1 - f_2 \]
Substituting the calculated values:
\[ \Delta f = 523.08 - 478.87 \approx 44.21 \, \text{Hz} \]

### Final Answer
The change in pitch of the note heard by the observer is approximately **44.21 Hz**.

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