An engine running at speed `(v)/(10)` sounds a whistle of frequency 600 Hz. A passenger in a train coming from the oppsite side at speed of `(v)/(15)` experiences this whistle to be of frequency f. If v is speed of sound in air and there is no wind. F is nearest to

Options

A. 711 Hz
B. 630 Hz
C. 580 Hz
D. 510 Hz

Explanation

To solve this problem, we will use the Doppler effect formula for sound waves. The formula for the apparent frequency \( f' \) is given by:

\[ f' = f \frac{v + v_0}{v - v_s} \]

Where:
- \( f' \) is the apparent frequency,
- \( f \) is the source frequency (600 Hz),
- \( v \) is the speed of sound in air,
- \( v_0 \) is the speed of the observer (train passenger),
- \( v_s \) is the speed of the source (engine).

### Step 1: Identify the given values
- Original frequency \( f = 600 \, \text{Hz} \)
- Speed of sound \( v \) (not specified numerically, but we will keep it as \( v \))
- Speed of the source \( v_s = \frac{v}{10} \)
- Speed of the observer \( v_0 = \frac{v}{15} \)

### Step 2: Substitute the values into the Doppler effect formula
Substituting the values into the formula, we have:

\[ f' = 600 \frac{v + \frac{v}{15}}{v - \frac{v}{10}} \]

### Step 3: Simplify the equation
First, simplify the numerator and denominator:

- **Numerator**:
\[ v + \frac{v}{15} = \frac{15v + v}{15} = \frac{16v}{15} \]

- **Denominator**:
\[ v - \frac{v}{10} = \frac{10v - v}{10} = \frac{9v}{10} \]

Now, substituting these back into the equation:

\[ f' = 600 \frac{\frac{16v}{15}}{\frac{9v}{10}} \]

### Step 4: Simplify further
Now we can simplify \( f' \):

\[ f' = 600 \cdot \frac{16v}{15} \cdot \frac{10}{9v} \]

The \( v \) cancels out:

\[ f' = 600 \cdot \frac{16 \cdot 10}{15 \cdot 9} \]

### Step 5: Calculate the numerical value
Now calculate the numerical value:

\[ f' = 600 \cdot \frac{160}{135} \]

Simplifying \( \frac{160}{135} \):

\[ \frac{160}{135} = \frac{32}{27} \]

Now substituting back:

\[ f' = 600 \cdot \frac{32}{27} \]

Calculating this gives:

\[ f' = \frac{19200}{27} \approx 711.11 \, \text{Hz} \]

### Conclusion
Thus, the apparent frequency \( f' \) experienced by the passenger is approximately \( 711 \, \text{Hz} \).

### Final Answer
The nearest frequency \( f \) is \( 711 \, \text{Hz} \).

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