Two open organ pipes vibrating I their fundamental mode produces 5 beats per second. Find the number of beats produced per second if the temperature of medium is made 2.56 times and the pipes still vibrate in their fundamental modes.

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript:

### Step 1: Understand the concept of beats
Beats occur when two sound waves of slightly different frequencies interfere with each other. The number of beats per second is given by the absolute difference in their frequencies:
\[ \text{Number of beats} = |f_1 - f_2| \]

### Step 2: Set up the initial condition
We know that the two open organ pipes produce 5 beats per second in their fundamental mode:
\[ |f_1 - f_2| = 5 \, \text{Hz} \]

### Step 3: Relate frequency to the speed of sound and pipe length
For an open organ pipe, the frequency is given by:
\[ f = \frac{v}{\lambda} \]
where \( v \) is the speed of sound and \( \lambda \) is the wavelength. For an open pipe, the wavelength is related to the length \( L \) of the pipe:
\[ \lambda = 2L \]
Thus, the frequency can be expressed as:
\[ f = \frac{v}{2L} \]

### Step 4: Express the frequencies of the two pipes
Let’s denote the velocities of sound in the two temperatures as \( v_1 \) and \( v_2 \), and the lengths of the pipes as \( L_1 \) and \( L_2 \). Therefore, we can write:
\[ f_1 = \frac{v_1}{2L_1} \]
\[ f_2 = \frac{v_2}{2L_2} \]

### Step 5: Set up the equations for the initial and final conditions
From the initial condition, we have:
\[ \frac{v_1}{2L_1} - \frac{v_2}{2L_2} = 5 \]

### Step 6: Analyze the effect of temperature change
When the temperature of the medium is increased to \( 2.56 \) times the original temperature, the speed of sound changes. The speed of sound is proportional to the square root of the temperature:
\[ v \propto \sqrt{T} \]
Thus, we have:
\[ v_2 = \sqrt{2.56} \cdot v_1 = 1.6 \cdot v_1 \]

### Step 7: Substitute the new speed of sound into the frequency equation
Now, substituting \( v_2 \) into the frequency difference equation:
\[ \frac{v_1}{2L_1} - \frac{1.6 v_1}{2L_2} = x \]
where \( x \) is the new number of beats per second.

### Step 8: Simplify the equation
Factoring out \( \frac{v_1}{2} \):
\[ \frac{v_1}{2} \left( \frac{1}{L_1} - \frac{1.6}{L_2} \right) = x \]

### Step 9: Relate the two equations
From the initial condition:
\[ \frac{v_1}{2} \left( \frac{1}{L_1} - \frac{1}{L_2} \right) = 5 \]

### Step 10: Find the new number of beats
Now, we can find the new number of beats \( x \) using the ratio of the two equations:
\[ \frac{\frac{1}{L_1} - \frac{1.6}{L_2}}{\frac{1}{L_1} - \frac{1}{L_2}} = \frac{x}{5} \]

### Step 11: Solve for \( x \)
This gives us:
\[ x = 5 \cdot \frac{1}{1.6} = 5 \cdot 0.625 = 3.125 \]

### Step 12: Final calculation
However, since we need to consider the increase in speed, we can directly calculate:
\[ x = 5 \cdot \sqrt{2.56} = 5 \cdot 1.6 = 8 \]

### Final Answer
The number of beats produced per second after increasing the temperature is:
\[ \boxed{8} \]

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