In a closed tube when air column is `20cm` it is in resonance with tuning fork A. When the length is increased by `2cm` then the air column is in resonance with tuning fork B. When A and B are sounded together they produce `8` beats per second . The frequencies of the tuning forks A and B are (in Hz)

Options

A. `40,44`
B. `88,30`
C. `80,88`
D. `44,40`

Explanation

To solve the problem step by step, we need to analyze the information given and apply the relevant physics concepts.

### Step 1: Understand the relationship between frequency and length of the air column
In a closed tube, the frequency of the sound produced is inversely proportional to the length of the air column. The fundamental frequency (first harmonic) for a closed tube can be expressed as:
\[ f = \frac{v}{4L} \]
where \( v \) is the speed of sound in air and \( L \) is the length of the air column.

### Step 2: Set up the equations for the two tuning forks
1. For tuning fork A, when the length of the air column is \( L_A = 20 \, \text{cm} = 0.2 \, \text{m} \):
\[ f_A = \frac{v}{4 \times 0.2} = \frac{v}{0.8} \]

2. For tuning fork B, when the length of the air column is increased by \( 2 \, \text{cm} \) (i.e., \( L_B = 22 \, \text{cm} = 0.22 \, \text{m} \)):
\[ f_B = \frac{v}{4 \times 0.22} = \frac{v}{0.88} \]

### Step 3: Relate the frequencies of the tuning forks
From the equations above, we can express the frequencies in terms of the speed of sound:
- \( f_A = \frac{v}{0.8} \)
- \( f_B = \frac{v}{0.88} \)

### Step 4: Use the beat frequency information
The problem states that when A and B are sounded together, they produce \( 8 \) beats per second. The beat frequency is given by the absolute difference between the two frequencies:
\[ |f_A - f_B| = 8 \]

### Step 5: Substitute the expressions for frequencies
Substituting the expressions for \( f_A \) and \( f_B \):
\[ \left| \frac{v}{0.8} - \frac{v}{0.88} \right| = 8 \]

### Step 6: Simplify the equation
To simplify, we can find a common denominator:
\[ \left| \frac{v \cdot 0.88 - v \cdot 0.8}{0.8 \cdot 0.88} \right| = 8 \]
\[ \left| \frac{v(0.88 - 0.8)}{0.8 \cdot 0.88} \right| = 8 \]
\[ \left| \frac{v(0.08)}{0.704} \right| = 8 \]

### Step 7: Solve for the speed of sound \( v \)
Now we can solve for \( v \):
\[ \frac{0.08v}{0.704} = 8 \]
\[ 0.08v = 8 \times 0.704 \]
\[ v = \frac{8 \times 0.704}{0.08} \]
\[ v = \frac{5.632}{0.08} \]
\[ v = 70.4 \, \text{m/s} \]

### Step 8: Calculate the frequencies \( f_A \) and \( f_B \)
Now we can find \( f_A \) and \( f_B \):
1. For \( f_A \):
\[ f_A = \frac{70.4}{0.8} = 88 \, \text{Hz} \]

2. For \( f_B \):
\[ f_B = \frac{70.4}{0.88} = 80 \, \text{Hz} \]

### Final Answer
The frequencies of the tuning forks A and B are:
- \( f_A = 88 \, \text{Hz} \)
- \( f_B = 80 \, \text{Hz} \)

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