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 find the frequency of the first tuning fork, we can follow these steps:

### Step 1: Define the frequency of the first tuning fork
Let the frequency of the first tuning fork be \( F_0 \).

### Step 2: Determine the frequency of the subsequent tuning forks
Since each tuning fork gives 4 beats per second with the preceding one, the frequency of the second tuning fork will be:
\[ F_1 = F_0 + 4 \]
The frequency of the third tuning fork will be:
\[ F_2 = F_0 + 8 \]
Continuing this pattern, the frequency of the \( n \)-th tuning fork can be expressed as:
\[ F_n = F_0 + 4(n-1) \]

### Step 3: Find the frequency of the 56th tuning fork
For the 56th tuning fork, we have:
\[ F_{56} = F_0 + 4(56 - 1) = F_0 + 220 \]

### Step 4: Use the octave relationship
According to the problem, the last fork (56th) is an octave higher than the first fork. This means:
\[ F_{56} = 2F_0 \]

### Step 5: Set up the equation
Now we can set up the equation using the expressions we have:
\[ F_0 + 220 = 2F_0 \]

### Step 6: Solve for \( F_0 \)
Rearranging the equation gives:
\[ 220 = 2F_0 - F_0 \]
\[ 220 = F_0 \]

### Conclusion
Thus, the frequency of the first tuning fork is:
\[ F_0 = 220 \text{ Hz} \]