A source of sound and an observer are approaching each other with the same speed, which is equal to `(1/10)` times the speed of sound. The apparent relative change in frequency of source is

Options

A. 22.2% increases
B. 22.2% decreases
C. 18.2% decreases
D. 18.2% increases

Answer

To solve the problem of the apparent relative change in frequency when a source of sound and an observer are approaching each other with the same speed, we can follow these steps:

### Step 1: Understand the given information
- Let the speed of sound be \( V \).
- The speed of both the source and the observer is \( \frac{V}{10} \).

### Step 2: Write the formula for apparent frequency
The formula for the apparent frequency \( f' \) when both the source and the observer are moving towards each other is given by:

\[ f' = f \left( \frac{V + V_o}{V - V_s} \right) \]

Where:
- \( f \) is the original frequency of the source.
- \( V_o \) is the speed of the observer.
- \( V_s \) is the speed of the source.

### Step 3: Substitute the values into the formula
Since both the observer and the source are moving with the same speed \( \frac{V}{10} \):

- \( V_o = \frac{V}{10} \)
- \( V_s = \frac{V}{10} \)

Substituting these values into the formula gives:

\[ f' = f \left( \frac{V + \frac{V}{10}}{V - \frac{V}{10}} \right) \]

### Step 4: Simplify the expression
Now simplify the fraction:

\[ f' = f \left( \frac{V + \frac{V}{10}}{V - \frac{V}{10}} \right) = f \left( \frac{V(1 + \frac{1}{10})}{V(1 - \frac{1}{10})} \right) \]

This simplifies to:

\[ f' = f \left( \frac{1 + \frac{1}{10}}{1 - \frac{1}{10}} \right) = f \left( \frac{\frac{11}{10}}{\frac{9}{10}} \right) = f \left( \frac{11}{9} \right) \]

### Step 5: Calculate the relative change in frequency
The relative change in frequency is given by:

\[ \text{Relative Change} = \frac{f' - f}{f} \times 100\% \]

Substituting \( f' = f \left( \frac{11}{9} \right) \):

\[ \text{Relative Change} = \frac{f \left( \frac{11}{9} \right) - f}{f} \times 100\% \]

This simplifies to:

\[ \text{Relative Change} = \left( \frac{\frac{11}{9} - 1}{1} \right) \times 100\% = \left( \frac{\frac{11 - 9}{9}}{1} \right) \times 100\% = \left( \frac{2}{9} \right) \times 100\% \]

Calculating this gives:

\[ \text{Relative Change} \approx 22.22\% \]

### Final Answer
The apparent relative change in frequency is approximately **22.22%**.

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