Two tuning forks `A and B` give `18 beats "in" 2 s`. A resonates with one end closed air column of `15 cm` long and `B` with both ends open column of `30.5` long. Calculate their frequencies.

Explanation

To solve the problem of finding the frequencies of the tuning forks A and B, we will follow these steps:

### Step 1: Understand the Beat Frequency
The problem states that two tuning forks A and B give 18 beats in 2 seconds. This means the beat frequency (the difference in frequencies of the two tuning forks) is:
\[ \text{Beat frequency} = \frac{18 \text{ beats}}{2 \text{ s}} = 9 \text{ Hz} \]
Thus, we have:
\[ |F_A - F_B| = 9 \text{ Hz} \]

### Step 2: Determine the Frequencies of the Tuning Forks
Tuning fork A resonates with a closed-end air column of length \(L_A = 15 \text{ cm} = 0.15 \text{ m}\). The frequency for a closed-end column is given by:
\[ F_A = \frac{V}{4L_A} \]
where \(V\) is the speed of sound in air (approximately \(343 \text{ m/s}\) at room temperature).

Tuning fork B resonates with an open-end air column of length \(L_B = 30.5 \text{ cm} = 0.305 \text{ m}\). The frequency for an open-end column is given by:
\[ F_B = \frac{V}{2L_B} \]

### Step 3: Substitute the Lengths into the Frequency Formulas
Substituting the lengths into the frequency formulas, we get:
\[ F_A = \frac{V}{4 \times 0.15} = \frac{V}{0.6} \]
\[ F_B = \frac{V}{2 \times 0.305} = \frac{V}{0.61} \]

### Step 4: Set Up the Equation Using the Beat Frequency
From the beat frequency, we know:
\[ |F_A - F_B| = 9 \]
Substituting the expressions for \(F_A\) and \(F_B\):
\[ \left|\frac{V}{0.6} - \frac{V}{0.61}\right| = 9 \]

### Step 5: Simplify the Equation
To simplify:
\[ \frac{V}{0.6} - \frac{V}{0.61} = 9 \]
Finding a common denominator:
\[ \frac{V \cdot 0.61 - V \cdot 0.6}{0.6 \cdot 0.61} = 9 \]
This simplifies to:
\[ \frac{V(0.61 - 0.6)}{0.6 \cdot 0.61} = 9 \]
\[ \frac{V(0.01)}{0.6 \cdot 0.61} = 9 \]

### Step 6: Solve for V
Cross-multiplying gives:
\[ V(0.01) = 9 \cdot (0.6 \cdot 0.61) \]
Calculating the right-hand side:
\[ 0.6 \cdot 0.61 = 0.366 \]
Thus:
\[ V(0.01) = 9 \cdot 0.366 \]
\[ V(0.01) = 3.294 \]
So:
\[ V = \frac{3.294}{0.01} = 329.4 \text{ m/s} \]

### Step 7: Calculate Frequencies
Now we can substitute \(V\) back into the equations for \(F_A\) and \(F_B\):
\[ F_A = \frac{329.4}{0.6} = 549 \text{ Hz} \]
\[ F_B = \frac{329.4}{0.61} \approx 540 \text{ Hz} \]

### Final Answer
The frequencies of the tuning forks are:
- \(F_A \approx 549 \text{ Hz}\)
- \(F_B \approx 540 \text{ Hz}\)