A fighter plane moving in a vertical loop with constant speed of radius `R` . The center of the loop is as a height `h` directly overhead of an observer standing on the ground. The observer receives maximum frequency of the sound produced by the plane when it is nearest to him . Find the speed of the plane. Velocity of sound in air is `upsilon` .

Explanation

To find the speed of the fighter plane moving in a vertical loop with radius \( R \), we can use the concept of the Doppler effect and the geometry of the situation. Here’s a step-by-step solution:

### Step 1: Understand the Geometry
The observer is standing on the ground at a distance \( h \) directly below the center of the loop. When the plane is at its nearest point to the observer, it is directly above the observer. The distance from the center of the loop to the observer is \( h \), and the radius of the loop is \( R \).

### Step 2: Identify the Relevant Distances
When the plane is at the nearest point, the distance from the plane to the observer (let's call this distance \( SO \)) can be calculated using the Pythagorean theorem:
\[ SO = \sqrt{h^2 + R^2} \]

### Step 3: Use the Doppler Effect
The observer receives the maximum frequency when the plane is moving directly towards him. According to the Doppler effect, the frequency heard by the observer is maximized when the source (the plane) is moving towards the observer. The time taken for the sound to travel from the plane to the observer must equal the time taken for the plane to move a small distance towards the observer.

### Step 4: Set Up the Time Equations
Let \( v_s \) be the speed of the plane and \( v \) be the speed of sound in air. The time taken for sound to travel from the plane to the observer is:
\[ t = \frac{SO}{v} \]
The time taken for the plane to move a distance \( \Delta s \) towards the observer is:
\[ t = \frac{\Delta s}{v_s} \]

### Step 5: Relate the Distances and Speeds
For maximum frequency, these two times must be equal:
\[ \frac{SO}{v} = \frac{\Delta s}{v_s} \]
Rearranging gives:
\[ v_s = \frac{\Delta s \cdot v}{SO} \]

### Step 6: Determine \( \Delta s \)
The distance \( \Delta s \) is the arc length the plane travels in a very small time interval. For small angles, this can be approximated as:
\[ \Delta s = R \cdot \theta \]
where \( \theta \) is the angle subtended at the center of the loop.

### Step 7: Substitute Back
Substituting \( \Delta s \) into the equation gives:
\[ v_s = \frac{R \cdot \theta \cdot v}{SO} \]
Now, substituting \( SO = \sqrt{h^2 + R^2} \):
\[ v_s = \frac{R \cdot \theta \cdot v}{\sqrt{h^2 + R^2}} \]

### Step 8: Find \( \theta \)
Using trigonometry, we find \( \theta \) using the cosine function:
\[ \cos \theta = \frac{R}{h} \implies \theta = \cos^{-1}\left(\frac{R}{h}\right) \]

### Step 9: Final Expression
Substituting \( \theta \) back into the equation gives:
\[ v_s = \frac{R \cdot \cos^{-1}\left(\frac{R}{h}\right) \cdot v}{\sqrt{h^2 + R^2}} \]

### Conclusion
The speed of the plane \( v_s \) can be expressed as:
\[ v_s = \frac{R \cdot v \cdot \cos^{-1}\left(\frac{R}{h}\right)}{\sqrt{h^2 + R^2}} \]