Fifty-six tuning forks are arranged in order of increasing frequencies so that each fork gives 4 beats per second with the next one. The last fork gives the octave of the first. Find the frequency of the first.

Options

A. 138 Hz
B. 144 Hz
C. 132 Hz
D. 276 Hz

Answer

To solve the problem step by step, let's denote the frequency of the first tuning fork as \( f \).

### Step 1: Define the frequencies of the tuning forks
The frequencies of the tuning forks can be expressed as follows:
- The frequency of the first tuning fork \( F_1 = f \)
- The frequency of the second tuning fork \( F_2 = f + 4 \)
- The frequency of the third tuning fork \( F_3 = f + 8 \)
- Continuing this pattern, the frequency of the \( n \)-th tuning fork can be expressed as:
\[ F_n = f + 4(n - 1) \]

### Step 2: Find the frequency of the last tuning fork
For the 56th tuning fork, we have:
\[ F_{56} = f + 4(56 - 1) = f + 4 \times 55 = f + 220 \]

### Step 3: Use the octave condition
The problem states that the last fork gives the octave of the first. This means:
\[ F_{56} = 2F_1 \]
Substituting the expressions we have:
\[ f + 220 = 2f \]

### Step 4: Solve for the frequency of the first tuning fork
Rearranging the equation:
\[ 220 = 2f - f \]
\[ 220 = f \]

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