A whistle producing sound waves of frequencies `9500 Hz` and above is approaching a stationary person with speed `vms^(-1)`. The velocity of sound in air is `300 ms^(-1)`. If the person can hear frequencies upto a maximum of `10,000 Hz`.The maximum value of `v` upto which he can hear whistle is

Options

A. (a)`15sqrt(2)m//s`
B. (b)`(15)/(sqrt(2))m//s`
C. (c)15m/s
D. (d)30m/s

Explanation

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

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

Where:
- \( f' \) is the observed frequency (10,000 Hz in this case),
- \( f \) is the source frequency (9,500 Hz),
- \( v \) is the speed of sound in air (300 m/s),
- \( v_s \) is the speed of the source (the whistle).

### Step 1: Write down the known values
- \( f' = 10,000 \, \text{Hz} \)
- \( f = 9,500 \, \text{Hz} \)
- \( v = 300 \, \text{m/s} \)

### Step 2: Substitute the known values into the Doppler effect formula
Using the formula, we substitute the known values:

\[ 10,000 = 9,500 \frac{300}{300 - v_s} \]

### Step 3: Rearrange the equation to solve for \( v_s \)
To isolate \( v_s \), we first multiply both sides by \( (300 - v_s) \):

\[ 10,000(300 - v_s) = 9,500 \times 300 \]

Expanding this gives:

\[ 3,000,000 - 10,000 v_s = 2,850,000 \]

### Step 4: Move terms involving \( v_s \) to one side
Now, we can move \( 10,000 v_s \) to the right side and \( 2,850,000 \) to the left side:

\[ 3,000,000 - 2,850,000 = 10,000 v_s \]

This simplifies to:

\[ 150,000 = 10,000 v_s \]

### Step 5: Solve for \( v_s \)
Now, we divide both sides by 10,000:

\[ v_s = \frac{150,000}{10,000} = 15 \, \text{m/s} \]

### Conclusion
The maximum speed \( v \) of the whistle, so that the person can still hear the sound, is **15 m/s**.