Two closed pipe produce 10 beats per second when emitting their fundamental nodes. If their length are in ratio of 25 : 26. Then their fundamental frequency in Hz , are

Answer

To solve the problem step by step, we need to find the fundamental frequencies of two closed pipes given the ratio of their lengths and the number of beats produced.

### Step 1: Understand the relationship between frequency and length
For closed pipes, the fundamental frequency (n) is inversely proportional to the length (L) of the pipe. The relationship can be expressed as:
\[ n = \frac{v}{4L} \]
where \( v \) is the speed of sound in air.

### Step 2: Set up the equations for the two pipes
Let the lengths of the two pipes be \( L_1 \) and \( L_2 \). According to the problem, the lengths are in the ratio:
\[ \frac{L_1}{L_2} = \frac{25}{26} \]
This implies:
\[ L_1 = 25k \quad \text{and} \quad L_2 = 26k \]
for some constant \( k \).

### Step 3: Write the frequencies in terms of \( k \)
Using the formula for frequency, we can express the frequencies of the two pipes as:
\[ n_1 = \frac{v}{4L_1} = \frac{v}{4 \times 25k} = \frac{v}{100k} \]
\[ n_2 = \frac{v}{4L_2} = \frac{v}{4 \times 26k} = \frac{v}{104k} \]

### Step 4: Use the information about beats
The number of beats per second is given as 10. The beat frequency is the absolute difference between the two frequencies:
\[ |n_1 - n_2| = 10 \]
Substituting the expressions for \( n_1 \) and \( n_2 \):
\[ \left| \frac{v}{100k} - \frac{v}{104k} \right| = 10 \]

### Step 5: Simplify the equation
To simplify, we can find a common denominator:
\[ \left| \frac{104v - 100v}{10400k} \right| = 10 \]
This simplifies to:
\[ \left| \frac{4v}{10400k} \right| = 10 \]
Thus:
\[ \frac{4v}{10400k} = 10 \]
Multiplying both sides by \( 10400k \):
\[ 4v = 104000k \]
Dividing both sides by 4:
\[ v = 26000k \]

### Step 6: Find the frequencies
Now, substituting \( v \) back into the equations for \( n_1 \) and \( n_2 \):
\[ n_1 = \frac{26000k}{100k} = 260 \, \text{Hz} \]
\[ n_2 = \frac{26000k}{104k} = 250 \, \text{Hz} \]

### Final Answer
The fundamental frequencies of the two pipes are:
- \( n_1 = 260 \, \text{Hz} \)
- \( n_2 = 250 \, \text{Hz} \)