A man standing on a platform hears the sound of frequency 604 Hz coming from a frequency 550 Hz from a train whistle moving towards the platform. If the velocity of sound is 330 m/s, then what is the speed of train?

Options

A. `30(m)/(s)`
B. `35(m)/(s)`
C. `40(m)/(s)`
D. `45(m)/(s)`

Explanation

To solve the problem, we will use the Doppler effect formula for sound waves when the source is moving towards a stationary observer. The formula for the apparent frequency (f') heard by the observer is given by:

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

Where:
- \( f' \) = apparent frequency (604 Hz)
- \( f \) = actual frequency of the source (550 Hz)
- \( v \) = speed of sound in air (330 m/s)
- \( v_0 \) = speed of the observer (0 m/s, since the man is stationary)
- \( v_s \) = speed of the source (train), which we need to find

### Step 1: Write down the known values
- \( f' = 604 \, \text{Hz} \)
- \( f = 550 \, \text{Hz} \)
- \( v = 330 \, \text{m/s} \)
- \( v_0 = 0 \, \text{m/s} \)

### Step 2: Substitute the known values into the Doppler effect formula
Since the observer is stationary, \( v_0 = 0 \). The formula simplifies to:

\[ 604 = \frac{330 + 0}{330 - v_s} \cdot 550 \]

### Step 3: Rearrange the equation to solve for \( v_s \)
First, multiply both sides by \( (330 - v_s) \):

\[ 604(330 - v_s) = 330 \cdot 550 \]

Now, expand the left side:

\[ 199320 - 604v_s = 181500 \]

### Step 4: Isolate \( v_s \)
Now, move \( 604v_s \) to the right side and \( 181500 \) to the left side:

\[ 199320 - 181500 = 604v_s \]

Calculate the left side:

\[ 17820 = 604v_s \]

### Step 5: Solve for \( v_s \)
Now, divide both sides by 604:

\[ v_s = \frac{17820}{604} \]

Calculating this gives:

\[ v_s \approx 29.5 \, \text{m/s} \]

### Step 6: Round to two significant figures
Rounding gives us:

\[ v_s \approx 30 \, \text{m/s} \]

### Final Answer
The speed of the train is approximately **30 m/s**.

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