A road passes at some distance from a standing man. A truck is coming on the road with some acceleration. The truck driver blows a whistle of frequency `500 Hz` when the line joining the truck and the man makes an angle `theta` with the road. The man hears a note having a note having a frequency of `600 Hz` when the truck is closest to him. Also the speed of truck has get doubled during this time. Find the value of `'theta'`.

Explanation

To solve the problem, we need to analyze the situation using the principles of sound waves and kinematics. Here’s a step-by-step solution:

### Step 1: Understand the scenario
A truck is moving towards a man while blowing a whistle of frequency \( f_0 = 500 \) Hz. The man hears a frequency \( f' = 600 \) Hz when the truck is closest to him. The truck's speed doubles during this time. We need to find the angle \( \theta \) between the line joining the truck and the man and the road.

### Step 2: Use the Doppler Effect formula
The frequency heard by the observer (the man) when the source (the truck) is moving towards him can be expressed using the Doppler Effect formula:

\[ f' = f_0 \frac{v}{v - v_s \cos \theta} \]

Where:
- \( f' = 600 \) Hz (frequency heard by the man)
- \( f_0 = 500 \) Hz (frequency of the whistle)
- \( v \) = speed of sound
- \( v_s \) = speed of the truck
- \( \theta \) = angle between the line joining the truck and the man and the road

### Step 3: Set up the equation
Substituting the known values into the Doppler Effect formula:

\[ 600 = 500 \frac{v}{v - v_s \cos \theta} \]

### Step 4: Rearranging the equation
Rearranging the equation gives:

\[ \frac{600}{500} = \frac{v}{v - v_s \cos \theta} \]

This simplifies to:

\[ \frac{6}{5} = \frac{v}{v - v_s \cos \theta} \]

Cross-multiplying gives:

\[ 6(v - v_s \cos \theta) = 5v \]

Expanding this results in:

\[ 6v - 6v_s \cos \theta = 5v \]

Thus, we have:

\[ v = 6v_s \cos \theta \]

### Step 5: Relate speed of the truck
From kinematics, we know that the final speed \( v_f \) of the truck is double the initial speed \( v_s \):

\[ v_f = 2v_s \]

Using the equation of motion for uniformly accelerated motion:

\[ v_f = v_s + a t \]

Where \( a \) is the acceleration and \( t \) is the time. Since \( v_f = 2v_s \), we can set up the equation:

\[ 2v_s = v_s + a t \implies a t = v_s \]

### Step 6: Distance traveled by the truck
The distance traveled by the truck while accelerating can be expressed as:

\[ d = v_s t + \frac{1}{2} a t^2 \]

Substituting \( a = \frac{v_s}{t} \):

\[ d = v_s t + \frac{1}{2} \left(\frac{v_s}{t}\right) t^2 = v_s t + \frac{1}{2} v_s t = \frac{3}{2} v_s t \]

### Step 7: Relate distance and angle
The distance \( d \) can also be expressed in terms of \( \theta \):

\[ d = L \cot \theta \]

Setting these equal gives:

\[ L \cot \theta = \frac{3}{2} v_s t \]

### Step 8: Find time in terms of \( \theta \)
The time taken for the sound to travel to the man is:

\[ t = \frac{L}{v \sin \theta} \]

### Step 9: Substitute \( t \) into the distance equation
Substituting this expression for \( t \) into the distance equation gives:

\[ L \cot \theta = \frac{3}{2} v_s \left(\frac{L}{v \sin \theta}\right) \]

### Step 10: Simplify and solve for \( \theta \)
Cancelling \( L \) from both sides and simplifying gives:

\[ \cot \theta = \frac{3}{2} \frac{v_s}{v \sin \theta} \]

Substituting \( v_s = \frac{2}{3} v \cos \theta \) into this equation leads to:

\[ \cot \theta = \frac{3}{2} \cdot \frac{2}{3} \cdot \frac{v \cos \theta}{v \sin \theta} \]

This simplifies to:

\[ \cot \theta = \cos \theta \]

### Step 11: Solve for \( \theta \)
This implies:

\[ \cot \theta = \cos \theta \implies \tan \theta = \frac{1}{\cos \theta} \]

This leads to:

\[ \theta = 45^\circ \]

### Final Answer
Thus, the value of \( \theta \) is \( 45^\circ \).

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