An observer moves on the same line on which two sources of sound of frequency `660 Hz` are present. The observer observes beat frequency of `10Hz`. If speed of sound is `300m//s` then speed of the observer is:

Options

A. `6m//s`
B. `3m//s`
C. `2.5m//s`
D. `4m//s`

Answer

To find the speed of the observer, we can follow these steps:

### Step 1: Understand the problem
We have two sound sources, both emitting sound at a frequency \( f = 660 \, \text{Hz} \). The observer is moving along the line connecting the two sources and hears a beat frequency of \( 10 \, \text{Hz} \). The speed of sound in air is given as \( v = 300 \, \text{m/s} \).

### Step 2: Determine the apparent frequencies
When the observer moves towards one source (let's call it \( S_1 \)), the apparent frequency \( f_1 \) can be calculated using the formula:
\[ f_1 = \left( \frac{v + v_0}{v} \right) f \]
where \( v_0 \) is the speed of the observer.

When the observer moves away from the other source (let's call it \( S_2 \)), the apparent frequency \( f_2 \) is given by:
\[ f_2 = \left( \frac{v - v_0}{v} \right) f \]

### Step 3: Set up the equation for beat frequency
The beat frequency \( f_b \) is the difference between the two apparent frequencies:
\[ f_b = |f_1 - f_2| = 10 \, \text{Hz} \]

Substituting the expressions for \( f_1 \) and \( f_2 \):
\[ \left| \left( \frac{v + v_0}{v} \right) f - \left( \frac{v - v_0}{v} \right) f \right| = 10 \]

### Step 4: Simplify the equation
Factoring out \( f \):
\[ \left| \left( \frac{(v + v_0) - (v - v_0)}{v} \right) f \right| = 10 \]
This simplifies to:
\[ \left| \frac{2v_0}{v} f \right| = 10 \]

### Step 5: Substitute known values
Substituting \( v = 300 \, \text{m/s} \) and \( f = 660 \, \text{Hz} \):
\[ \frac{2v_0}{300} \cdot 660 = 10 \]

### Step 6: Solve for \( v_0 \)
Rearranging the equation gives:
\[ 2v_0 \cdot 660 = 10 \cdot 300 \]
\[ 2v_0 \cdot 660 = 3000 \]
\[ v_0 \cdot 660 = 1500 \]
\[ v_0 = \frac{1500}{660} \approx 2.27 \, \text{m/s} \]

### Final Answer
The speed of the observer \( v_0 \) is approximately \( 2.27 \, \text{m/s} \).

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