A bat emitting an ultrasonic wave of frequecy `4.5 xx 10^4` Hz flies at a speed of `6ms^(-1)` between two parallel walls. Find the two frequecies heard by the bat and the beat frequecy between the two. The speed of sound is `330ms^(-1)`

Explanation

To solve the problem, we will use the concept of the Doppler effect to find the two frequencies heard by the bat and then calculate the beat frequency between them.

### Step 1: Identify the given data
- Frequency emitted by the bat (f₀) = \(4.5 \times 10^4\) Hz
- Speed of the bat (vₛ) = 6 m/s
- Speed of sound (v) = 330 m/s

### Step 2: Calculate the frequency heard by wall 1 (f₁)
When the bat is flying towards wall 1, the wall acts as an observer. We can use the Doppler effect formula:

\[ f_1 = f_0 \times \frac{v + v_o}{v - v_s} \]

Where:
- \(v_o\) (velocity of observer) = 0 (since the wall is stationary)
- \(v_s\) (velocity of source) = 6 m/s (since the bat is moving towards the wall)

Substituting the values:

\[ f_1 = 4.5 \times 10^4 \times \frac{330 + 0}{330 - 6} \]

Calculating the denominator:

\[ 330 - 6 = 324 \]

Now substituting back:

\[ f_1 = 4.5 \times 10^4 \times \frac{330}{324} \]

Calculating \(f_1\):

\[ f_1 \approx 4.5 \times 10^4 \times 1.0185 \approx 4.586 \times 10^4 \text{ Hz} \]

### Step 3: Calculate the frequency heard by wall 2 (f₂)
When the bat is flying away from wall 2, the wall again acts as an observer. The formula remains the same, but now the bat is moving away from the wall:

\[ f_2 = f_0 \times \frac{v + v_o}{v + v_s} \]

Where:
- \(v_o\) = 0 (the wall is stationary)
- \(v_s\) = 6 m/s (the bat is moving away from the wall)

Substituting the values:

\[ f_2 = 4.5 \times 10^4 \times \frac{330 + 0}{330 + 6} \]

Calculating the denominator:

\[ 330 + 6 = 336 \]

Now substituting back:

\[ f_2 = 4.5 \times 10^4 \times \frac{330}{336} \]

Calculating \(f_2\):

\[ f_2 \approx 4.5 \times 10^4 \times 0.9821 \approx 4.409 \times 10^4 \text{ Hz} \]

### Step 4: Calculate the beat frequency (n)
The beat frequency is given by the absolute difference between the two frequencies:

\[ n = |f_1 - f_2| \]

Substituting the values:

\[ n = |4.586 \times 10^4 - 4.409 \times 10^4| \]

Calculating \(n\):

\[ n \approx |4.586 - 4.409| \times 10^4 \approx 0.177 \times 10^4 \approx 1770 \text{ Hz} \]

### Final Answers
- Frequency heard by wall 1 (f₁) ≈ \(4.586 \times 10^4\) Hz
- Frequency heard by wall 2 (f₂) ≈ \(4.409 \times 10^4\) Hz
- Beat frequency (n) ≈ 1770 Hz