Bus moving with speed v forwards a stationary wall. it produces sound of frequency `f = 420 Hz. The heard frequency of reflected sound from wall by driver is `490 Hz`. Calculate the speed v of bus. The velocity of sound in air is `330 m/s`.

Options

A. `61 Km/s`
B. `71 Km/s
C. `81 Km/s`
D. `91 Km/s`

Answer

To solve the problem, we need to apply the Doppler effect for sound. The bus is moving towards a stationary wall, and it emits a sound of frequency \( f = 420 \, \text{Hz} \). The frequency of the reflected sound heard by the driver is \( f' = 490 \, \text{Hz} \). The speed of sound in air is given as \( c = 330 \, \text{m/s} \).

### Step-by-Step Solution:

1. **Identify the frequencies and velocities:**
- Frequency emitted by the bus: \( f = 420 \, \text{Hz} \)
- Frequency heard by the driver after reflection: \( f' = 490 \, \text{Hz} \)
- Speed of sound: \( c = 330 \, \text{m/s} \)
- Speed of the bus: \( v \) (unknown)

2. **Apply the Doppler effect:**
When the bus emits sound, the frequency heard by the wall (which is stationary) can be calculated using the formula for the Doppler effect:
\[ f' = f \cdot \frac{c}{c - v} \]
Here, \( f' \) is the frequency heard by the wall, \( f \) is the frequency emitted by the bus, \( c \) is the speed of sound, and \( v \) is the speed of the bus.

3. **Calculate the frequency reflected back to the bus:**
The wall reflects the sound back to the bus. The frequency heard by the bus after reflection is given by:
\[ f'' = f' \cdot \frac{c + v}{c} \]
Substituting \( f' \) from the previous step:
\[ f'' = \left( f \cdot \frac{c}{c - v} \right) \cdot \frac{c + v}{c} \]

4. **Set up the equation using the known frequencies:**
We know that \( f'' = 490 \, \text{Hz} \):
\[ 490 = 420 \cdot \frac{c + v}{c - v} \]

5. **Substitute the values and simplify:**
Substitute \( c = 330 \, \text{m/s} \):
\[ 490 = 420 \cdot \frac{330 + v}{330 - v} \]
Rearranging gives:
\[ \frac{330 + v}{330 - v} = \frac{490}{420} \]
Simplifying \( \frac{490}{420} \) gives:
\[ \frac{490}{420} = \frac{49}{42} = \frac{7}{6} \]

6. **Cross-multiply to solve for \( v \):**
\[ 6(330 + v) = 7(330 - v) \]
Expanding both sides:
\[ 1980 + 6v = 2310 - 7v \]
Combine like terms:
\[ 6v + 7v = 2310 - 1980 \]
\[ 13v = 330 \]
\[ v = \frac{330}{13} \approx 25.38 \, \text{m/s} \]

### Final Answer:
The speed of the bus \( v \) is approximately \( 25.38 \, \text{m/s} \).