The frequency of the sound of a car horn as recorded by an observer towards whom the car is moving differs from the frequency of the horn by `10%`. Assuming the velocity of sound in air to be `330ms^(-1)`, the velocity of the car is

Options

A. `36.7ms^(-1)`
B. `40ms^(-1)`
C. `30ms^(-1)`
D. `33ms^(-1)`

Explanation

To solve the problem, we will use the Doppler effect formula for sound. The apparent frequency \( n' \) heard by an observer when the source of sound is moving towards the observer is given by:

\[ n' = n \left( \frac{v + v_o}{v - v_s} \right) \]

Where:
- \( n' \) is the apparent frequency,
- \( n \) is the actual frequency,
- \( v \) is the speed of sound in air,
- \( v_o \) is the speed of the observer (0 in this case since the observer is at rest),
- \( v_s \) is the speed of the source (the car).

Given that the apparent frequency differs from the actual frequency by 10%, we can express this as:

\[ n' = 1.1n \]

Substituting this into the Doppler effect formula, we have:

\[ 1.1n = n \left( \frac{v + 0}{v - v_s} \right) \]

Since \( n \) is common on both sides, we can cancel it out (assuming \( n \neq 0 \)):

\[ 1.1 = \frac{v}{v - v_s} \]

Now, substituting the known values:
- \( v = 330 \, \text{m/s} \)

We can rewrite the equation:

\[ 1.1 = \frac{330}{330 - v_s} \]

Cross-multiplying gives:

\[ 1.1(330 - v_s) = 330 \]

Expanding this:

\[ 363 - 1.1v_s = 330 \]

Now, isolating \( v_s \):

\[ 363 - 330 = 1.1v_s \]

\[ 33 = 1.1v_s \]

Dividing both sides by 1.1:

\[ v_s = \frac{33}{1.1} = 30 \, \text{m/s} \]

Thus, the velocity of the car is \( 30 \, \text{m/s} \).

### Summary of Steps:
1. Write down the Doppler effect formula for sound.
2. Express the relationship between the apparent frequency and actual frequency.
3. Substitute the known values into the formula.
4. Cancel out the common terms and simplify the equation.
5. Solve for the speed of the car.