An object producing a pitch of 100 h=Hz approaches a stationary person in a straight line with a velocity of 200 m/s. Velocity of sound is 300 m/s. The person will note a change in frequency, as the object flies past him equal to

Options

A. 1440 Hz
B. 240 Hz
C. 1200 Hz
D. 960 Hz

Answer

To solve the problem of finding the change in frequency as an object producing a pitch of 100 Hz approaches and then moves away from a stationary observer, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Frequency of the source (f) = 100 Hz
- Velocity of the source (vs) = 200 m/s (approaching)
- Velocity of sound (v) = 300 m/s

2. **Calculate the Apparent Frequency when the Source Approaches the Observer:**
- The formula for the apparent frequency when the source is approaching the observer is given by:
\[ f_1 = \frac{v}{v - v_s} \cdot f \]
- Substituting the values:
\[ f_1 = \frac{300}{300 - 200} \cdot 100 = \frac{300}{100} \cdot 100 = 3 \cdot 100 = 300 \text{ Hz} \]

3. **Calculate the Apparent Frequency when the Source Moves Away from the Observer:**
- The formula for the apparent frequency when the source is moving away from the observer is given by:
\[ f_2 = \frac{v}{v + v_s} \cdot f \]
- Substituting the values:
\[ f_2 = \frac{300}{300 + 200} \cdot 100 = \frac{300}{500} \cdot 100 = 0.6 \cdot 100 = 60 \text{ Hz} \]

4. **Calculate the Change in Frequency:**
- The change in frequency (Δf) as the object flies past the observer is given by:
\[ \Delta f = f_1 - f_2 \]
- Substituting the values:
\[ \Delta f = 300 - 60 = 240 \text{ Hz} \]

5. **Final Answer:**
- The change in frequency noted by the person is **240 Hz**.