A military band was marching on a street in the same direction as the normal traffic flow. A motorist was approching the band from behind in his car just at the moment when they were playing the musical note `A` (frequency `440 Hz`) in unison. The same band' performance was also simultaneously being broad live on the ratio from a stationary microphone which was situated on the sidewalk just ahead of the band. The motorist noticed that the combination of the direct sound coming from the band ahead of the band brodcast produced discernible beats. Timing the beats, he counted `4` beats in `3`seconds. He noted from the speedmeter that his car was travelling at `18` kilometers per hour.
Calculate the speed in meters per seconds at which the band was marching assuming that the speed of sound on that day was `330 ms^(-1)`. (It was a very cold day)

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert the speed of the motorist from km/h to m/s
The motorist's speed is given as 18 km/h. To convert this to meters per second (m/s), we use the conversion factor:
\[ \text{Speed in m/s} = \frac{\text{Speed in km/h} \times 1000}{3600} \]
Calculating this gives:
\[ \text{Speed of motorist} = \frac{18 \times 1000}{3600} = 5 \text{ m/s} \]

### Step 2: Determine the beat frequency
The beat frequency is given as 4 beats in 3 seconds. To find the beat frequency in Hz (beats per second), we calculate:
\[ \text{Beat frequency} = \frac{4 \text{ beats}}{3 \text{ seconds}} = \frac{4}{3} \text{ Hz} \]

### Step 3: Set up the equations for apparent frequencies
1. **Apparent frequency at the stationary microphone (Fm')**:
- Since the microphone is stationary, the velocity of the observer (Vo) is 0, and the velocity of the source (Vs) is the speed of the band (V0):
\[ Fm' = f_0 \cdot \frac{v}{v - V_0} = 440 \cdot \frac{330}{330 - V_0} \]

2. **Apparent frequency at the car (Fc')**:
- The motorist is moving towards the band, so:
\[ Fc' = f_0 \cdot \frac{v + V_c}{v + V_0} = 440 \cdot \frac{330 + 5}{330 + V_0} = 440 \cdot \frac{335}{330 + V_0} \]

### Step 4: Relate the beat frequency to the apparent frequencies
The beat frequency is the difference between the two apparent frequencies:
\[ \text{Beat frequency} = |Fc' - Fm'| = \frac{4}{3} \]
Substituting the expressions for Fc' and Fm':
\[ \left| 440 \cdot \frac{335}{330 + V_0} - 440 \cdot \frac{330}{330 - V_0} \right| = \frac{4}{3} \]

### Step 5: Simplify and solve for V0
Factor out 440:
\[ 440 \left| \frac{335}{330 + V_0} - \frac{330}{330 - V_0} \right| = \frac{4}{3} \]
Dividing both sides by 440:
\[ \left| \frac{335}{330 + V_0} - \frac{330}{330 - V_0} \right| = \frac{4}{1320} \]
Now, we can solve this equation for V0. This involves finding a common denominator and simplifying the expression.

### Step 6: Solve the resulting equation
After solving, we will find two possible values for V0.

### Final Result
The speed of the band (V0) is approximately 1.985 m/s or 2.98 m/s.

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