An observer moves towards a stationary source of sound with a speed `1/5`th of the speed of sound. The wavelength and frequency of the source emited are `lamda` and f, respectively. The apparent frequency and wavelength recorded by the observer are, respectively-
(a) `f,1.2 lambda` (b) `0.8f,0.8lambda` (c) `1.2f,1.2lambda` (d) `1.2f,lambda`

Options

A. `f,1.2 lambda`
B. `0.8f,0.8lambda`
C. `1.2f,1.2lambda`
D. `1.2f,lambda`

Answer

To solve the problem step-by-step, we will apply the principles of the Doppler effect for sound waves.

### Step 1: Understand the given parameters
- The speed of sound is denoted as \( V \).
- The speed of the observer is \( \frac{1}{5} \) of the speed of sound, which can be expressed as:
\[ V_{\text{observer}} = \frac{V}{5} \]
- The frequency of the source is \( f \) and the wavelength is \( \lambda \).

### Step 2: Determine the apparent frequency using the Doppler effect formula
The formula for the apparent frequency \( f' \) when the observer is moving towards a stationary source is given by:
\[ f' = f \cdot \frac{V + V_{\text{observer}}}{V} \]
Since the source is stationary, its velocity is 0. Therefore, we can substitute the values into the formula:
\[ f' = f \cdot \frac{V + \frac{V}{5}}{V} \]

### Step 3: Simplify the expression for apparent frequency
Now, simplify the expression:
\[ f' = f \cdot \frac{V + \frac{V}{5}}{V} = f \cdot \frac{V(1 + \frac{1}{5})}{V} = f \cdot (1 + \frac{1}{5}) = f \cdot \frac{6}{5} \]
This simplifies to:
\[ f' = 1.2f \]

### Step 4: Determine the apparent wavelength
The wavelength is related to frequency and speed of sound by the equation:
\[ \lambda' = \frac{V}{f'} \]
Since the speed of sound \( V \) remains constant and the frequency has changed, we can express the new wavelength as:
\[ \lambda' = \frac{V}{1.2f} \]
Now, since the original wavelength is given by:
\[ \lambda = \frac{V}{f} \]
We can relate the new wavelength to the original wavelength:
\[ \lambda' = \frac{V}{1.2f} = \frac{1}{1.2} \cdot \frac{V}{f} = \frac{1}{1.2} \cdot \lambda \]
This simplifies to:
\[ \lambda' = 0.833\lambda \quad (\text{approximately } 0.8\lambda) \]

### Conclusion
Thus, the apparent frequency and wavelength recorded by the observer are:
- Apparent frequency: \( 1.2f \)
- Apparent wavelength: \( 0.8\lambda \)

The correct answer is option (d): \( 1.2f, \lambda \).