A set of `56 `tuning forks is arranged in a sequence of increasing frequencies . If each fork gives `4 beats//s` with the preceding one and the last fork is found to be an octave higher of the first , find the frequency of the first fork.

Explanation

To solve the problem, we need to find the frequency of the first tuning fork given the conditions provided. Let's break down the solution step by step.

### Step 1: Understand the relationship between frequencies
We know that the last tuning fork is an octave higher than the first. This means that if the frequency of the first fork is \( f_1 \), then the frequency of the last fork \( f_{56} \) can be expressed as:
\[ f_{56} = 2 f_1 \]

### Step 2: Determine the frequency difference between consecutive forks
Each tuning fork produces 4 beats per second with the preceding fork. The beat frequency is given by the absolute difference in frequencies of the two forks. Therefore, the difference between the frequencies of consecutive forks is:
\[ f_{n} - f_{n-1} = 4 \text{ Hz} \]
This means that if the frequency of the first fork is \( f_1 \), the frequency of the second fork \( f_2 \) will be:
\[ f_2 = f_1 + 4 \]
The frequency of the third fork \( f_3 \) will be:
\[ f_3 = f_1 + 2 \times 4 = f_1 + 8 \]
Continuing this pattern, the frequency of the \( n^{th} \) fork can be expressed as:
\[ f_n = f_1 + (n-1) \times 4 \]

### Step 3: Write the expression for the last fork
For the 56th fork, we can write:
\[ f_{56} = f_1 + (56-1) \times 4 = f_1 + 220 \]

### Step 4: Set up the equation
From Step 1, we have:
\[ f_{56} = 2 f_1 \]
From Step 3, we have:
\[ f_{56} = f_1 + 220 \]
Now we can set these two expressions for \( f_{56} \) equal to each other:
\[ 2 f_1 = f_1 + 220 \]

### Step 5: Solve for \( f_1 \)
Subtract \( f_1 \) from both sides:
\[ 2 f_1 - f_1 = 220 \]
This simplifies to:
\[ f_1 = 220 \text{ Hz} \]

### Conclusion
The frequency of the first tuning fork is:
\[ \boxed{220 \text{ Hz}} \]