A stationary source of sound is emitting sound of frequency `500Hz`. Two observers `A` and `B` lying on the same line as the source, observe frequencies `480Hz` and `530 Hz` respectively. The velocity of `A` and `B` respectively are (in `m//s`), speed of sound `=300m//s`.

Options

A. `12,18`
B. `8,18`
C. `18,8`
D. `18,12`

Answer

To solve the problem, we will use the Doppler effect formula for sound. The frequency observed by an observer moving relative to a stationary source can be calculated using the following formula:

\[ f' = \frac{v + v_o}{v} f \]

Where:
- \( f' \) is the observed frequency,
- \( v \) is the speed of sound,
- \( v_o \) is the velocity of the observer (positive if moving towards the source, negative if moving away),
- \( f \) is the frequency of the source.

### Step 1: Calculate the velocity of observer A

Given:
- Frequency of source, \( f = 500 \, \text{Hz} \)
- Observed frequency by A, \( f_A = 480 \, \text{Hz} \)
- Speed of sound, \( v = 300 \, \text{m/s} \)

Since observer A is moving away from the source, we will take \( v_o \) as negative.

Using the formula:
\[ f_A = \frac{v - v_A}{v} f \]

Substituting the known values:
\[ 480 = \frac{300 - v_A}{300} \times 500 \]

### Step 2: Rearranging the equation

Multiply both sides by 300:
\[ 480 \times 300 = (300 - v_A) \times 500 \]

Now calculate \( 480 \times 300 \):
\[ 144000 = (300 - v_A) \times 500 \]

### Step 3: Solve for \( 300 - v_A \)

Divide both sides by 500:
\[ \frac{144000}{500} = 300 - v_A \]
\[ 288 = 300 - v_A \]

### Step 4: Isolate \( v_A \)

Rearranging gives:
\[ v_A = 300 - 288 = 12 \, \text{m/s} \]

### Step 5: Calculate the velocity of observer B

Now, for observer B who is moving towards the source, we will take \( v_o \) as positive.

Using the formula:
\[ f_B = \frac{v + v_B}{v} f \]

Substituting the known values:
\[ 530 = \frac{300 + v_B}{300} \times 500 \]

### Step 6: Rearranging the equation for B

Multiply both sides by 300:
\[ 530 \times 300 = (300 + v_B) \times 500 \]

Calculating \( 530 \times 300 \):
\[ 159000 = (300 + v_B) \times 500 \]

### Step 7: Solve for \( 300 + v_B \)

Divide both sides by 500:
\[ \frac{159000}{500} = 300 + v_B \]
\[ 318 = 300 + v_B \]

### Step 8: Isolate \( v_B \)

Rearranging gives:
\[ v_B = 318 - 300 = 18 \, \text{m/s} \]

### Final Answer

The velocities of observers A and B are:
- \( v_A = 12 \, \text{m/s} \)
- \( v_B = 18 \, \text{m/s} \)