the frequency changes by 10 % as the source approaches a stationary observer with constant speed `v_(s)`. What would be the percentage change in frequency as the sources reaccedes the observer with the same speed ? Given , that `v_(s)lt lt v `(v= speed pf sound in air )

Options

A. 14. 3 %
B. 0.2
C. 16 . 7 %
D. 0.1

Answer

To solve the problem, we will use the principles of the Doppler effect. We are given that the frequency changes by 10% when the source approaches a stationary observer. We need to find the percentage change in frequency when the source recedes from the observer with the same speed.

### Step-by-Step Solution:

1. **Understanding the Doppler Effect**:
When a source of sound approaches a stationary observer, the observed frequency increases. Conversely, when the source recedes, the observed frequency decreases.

2. **Given Information**:
- The original frequency is \( f_0 \).
- The frequency increases by 10% when the source approaches, so the new frequency \( f_1 \) when the source approaches is:
\[ f_1 = 1.1 f_0 \]

3. **Using the Doppler Effect Formula for Approaching Source**:
The formula for the apparent frequency when the source is moving towards the observer is:
\[ f' = f_0 \frac{v}{v - v_s} \]
where \( v \) is the speed of sound, and \( v_s \) is the speed of the source.

4. **Setting Up the Equation for Approaching Source**:
From the above formula, we can set up the equation:
\[ 1.1 f_0 = f_0 \frac{v}{v - v_s} \]
Dividing both sides by \( f_0 \) and simplifying gives:
\[ 1.1 = \frac{v}{v - v_s} \]

5. **Solving for \( v_s \)**:
Rearranging the equation:
\[ 1.1(v - v_s) = v \]
\[ 1.1v - 1.1v_s = v \]
\[ 0.1v = 1.1v_s \]
\[ v_s = \frac{0.1v}{1.1} = \frac{v}{11} \]

6. **Finding the Frequency When the Source Recedes**:
Now, we use the Doppler effect formula for when the source is receding:
\[ f' = f_0 \frac{v}{v + v_s} \]
Substituting \( v_s = \frac{v}{11} \):
\[ f_2 = f_0 \frac{v}{v + \frac{v}{11}} = f_0 \frac{v}{\frac{11v + v}{11}} = f_0 \frac{v}{\frac{12v}{11}} = f_0 \frac{11}{12} \]

7. **Calculating the Percentage Change in Frequency**:
The change in frequency when the source recedes is:
\[ \Delta f = f_2 - f_0 = \left(\frac{11}{12} f_0 - f_0\right) = \left(\frac{11 - 12}{12}\right) f_0 = -\frac{1}{12} f_0 \]
The percentage change is given by:
\[ \text{Percentage Change} = \left(\frac{\Delta f}{f_0}\right) \times 100 = \left(-\frac{1}{12}\right) \times 100 = -8.33\% \]

### Final Answer:
The percentage change in frequency as the source recedes is approximately **8.33% decrease**.