A car has two horns having a difference in frequency of 180 Hz. The car is approaching a stationary observer with a speed of `60 ms^(-1)`. Calculate the difference in frequencies of the notes as heard by the observer, If velocity of sound in air is `330 ms^(-1)`.

Explanation

To solve the problem step by step, we will use the Doppler effect formula for sound. The Doppler effect describes the change in frequency (or wavelength) of a wave in relation to an observer who is moving relative to the wave source.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Frequency difference between the two horns, \( \Delta f = 180 \, \text{Hz} \)
- Speed of the car (source), \( v_s = 60 \, \text{m/s} \)
- Speed of sound in air, \( v = 330 \, \text{m/s} \)

2. **Use the Doppler Effect Formula:**
The formula to calculate the observed frequency difference when the source is moving towards a stationary observer is given by:
\[ \Delta f' = \Delta f \left( \frac{v}{v - v_s} \right) \]
where:
- \( \Delta f' \) is the observed frequency difference,
- \( v \) is the speed of sound,
- \( v_s \) is the speed of the source.

3. **Substitute the Values:**
Plugging in the values we have:
\[ \Delta f' = 180 \left( \frac{330}{330 - 60} \right) \]

4. **Calculate the Denominator:**
First, calculate \( 330 - 60 \):
\[ 330 - 60 = 270 \]

5. **Calculate the Fraction:**
Now calculate the fraction:
\[ \frac{330}{270} \]

6. **Simplify the Fraction:**
To simplify \( \frac{330}{270} \):
- Divide both numerator and denominator by 30:
\[ \frac{330 \div 30}{270 \div 30} = \frac{11}{9} \]

7. **Calculate the Observed Frequency Difference:**
Now substitute back into the equation:
\[ \Delta f' = 180 \times \frac{11}{9} \]

8. **Perform the Multiplication:**
Calculate \( 180 \times \frac{11}{9} \):
\[ \Delta f' = 180 \times 1.2222 \approx 220 \, \text{Hz} \]

9. **Final Result:**
The difference in frequencies of the notes as heard by the observer is:
\[ \Delta f' \approx 220 \, \text{Hz} \]

### Summary:
The observed difference in frequencies of the notes as heard by the observer is approximately **220 Hz**.

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