A whistle of frequency `f_0` = 1300 Hz is dropped from a height H = 505 m above the ground. At the same time, a detector is projected upwards with velocity `v = 50 ms^( -1)` along the same line. If the velocity of sound is `c = 300ms^(-1)` find the frequency detected by the detector after t = 5s.

Options

A. 500 Hz
B. 700 Hz
C. 1500 Hz
D. 2500 Hz

Answer

To solve the problem, we need to calculate the frequency detected by the detector after 5 seconds when the whistle is dropped from a height and the detector is moving upwards. We will use the Doppler effect formula for sound.

### Step 1: Determine the distance traveled by the whistle (source)
The whistle is dropped from a height \( H = 505 \, \text{m} \) with an initial velocity \( u = 0 \, \text{m/s} \). The distance traveled by the whistle after time \( t = 5 \, \text{s} \) can be calculated using the formula for distance under constant acceleration (gravity in this case).

\[ s = ut + \frac{1}{2} g t^2 \]
Where:
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( t = 5 \, \text{s} \)

Substituting the values:
\[ s = 0 \cdot 5 + \frac{1}{2} \cdot 10 \cdot (5)^2 = 0 + \frac{1}{2} \cdot 10 \cdot 25 = 125 \, \text{m} \]

### Step 2: Determine the position of the whistle after 5 seconds
The height of the whistle after 5 seconds will be:
\[ H' = H - s = 505 \, \text{m} - 125 \, \text{m} = 380 \, \text{m} \]

### Step 3: Determine the distance traveled by the detector (observer)
The detector is projected upwards with an initial velocity \( v_0 = 50 \, \text{m/s} \). The distance traveled by the detector after time \( t = 5 \, \text{s} \) is given by:
\[ s_o = v_0 t - \frac{1}{2} g t^2 \]
Substituting the values:
\[ s_o = 50 \cdot 5 - \frac{1}{2} \cdot 10 \cdot (5)^2 = 250 - 125 = 125 \, \text{m} \]

### Step 4: Determine the height of the detector after 5 seconds
The height of the detector after 5 seconds will be:
\[ H_d = 0 + s_o = 0 + 125 \, \text{m} = 125 \, \text{m} \]

### Step 5: Calculate the relative positions of the source and observer
After 5 seconds, the positions of the whistle and the detector are:
- Whistle: \( 380 \, \text{m} \) above the ground
- Detector: \( 125 \, \text{m} \) above the ground

### Step 6: Calculate the velocities of the source and observer
1. **Velocity of the source (whistle)** after 5 seconds:
\[ v_s = u + gt = 0 + 10 \cdot 5 = 50 \, \text{m/s} \, \text{(downward)} \]

2. **Velocity of the observer (detector)** after 5 seconds:
\[ v_o = v_0 - gt = 50 - 10 \cdot 5 = 0 \, \text{m/s} \, \text{(at the peak)} \]

### Step 7: Apply the Doppler effect formula
The frequency detected by the observer can be calculated using the Doppler effect formula:
\[ f' = f_0 \cdot \frac{c + v_o}{c - v_s} \]
Where:
- \( f_0 = 1300 \, \text{Hz} \)
- \( c = 300 \, \text{m/s} \)
- \( v_o = 0 \, \text{m/s} \)
- \( v_s = 50 \, \text{m/s} \)

Substituting the values:
\[ f' = 1300 \cdot \frac{300 + 0}{300 - 50} = 1300 \cdot \frac{300}{250} \]
Calculating:
\[ f' = 1300 \cdot 1.2 = 1560 \, \text{Hz} \]

### Final Answer
The frequency detected by the detector after 5 seconds is \( 1560 \, \text{Hz} \).

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