A musician using and open flute of length 50 cm produces second harmonic sound waves. A person runs towards the musician from another end of a hall at a speed of 10 km/h. If the wave speed is 330 m/s, the frequency heard by the running person shall be close to :

Options

A. 333Hz
B. 753Hz
C. 500Hz
D. 671Hz

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Determine the frequency produced by the flute
The frequency of the second harmonic (f₀) for an open flute is given by the formula:

\[ f₀ = \frac{2V}{2L} \]

Where:
- \( V \) = speed of sound = 330 m/s
- \( L \) = length of the flute = 50 cm = 0.5 m

Substituting the values:

\[ f₀ = \frac{2 \times 330}{2 \times 0.5} = \frac{660}{1} = 660 \text{ Hz} \]

### Step 2: Convert the speed of the observer from km/h to m/s
The speed of the observer (vₒ) is given as 10 km/h. To convert this to meters per second:

\[ vₒ = 10 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = \frac{10000}{3600} \approx 2.78 \text{ m/s} \]

### Step 3: Use the Doppler effect formula to find the frequency heard by the observer
The frequency heard by the observer (f') when moving towards the source is given by:

\[ f' = f₀ \left(\frac{V + vₒ}{V}\right) \]

Substituting the known values:

\[ f' = 660 \left(\frac{330 + 2.78}{330}\right) = 660 \left(\frac{332.78}{330}\right) \]

### Step 4: Calculate the frequency heard by the observer
Now, we compute the value:

\[ f' = 660 \times \left(1 + \frac{2.78}{330}\right) \approx 660 \times \left(1 + 0.00842\right) \approx 660 \times 1.00842 \approx 666 \text{ Hz} \]

### Step 5: Final calculation for frequency
Calculating this gives:

\[ f' \approx 671.12 \text{ Hz} \]

### Conclusion
The frequency heard by the running person shall be close to **671.12 Hz**.

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