56 tuning forks are so arranged in series that each fork give 4 beats per sec with the previous one. The frequency of the last fork is 3 times that of the first. The frequency of the first fork is

Options

A. 110
B. 56
C. 60
D. 52

Explanation

To solve the problem step by step, we will denote the frequency of the first tuning fork as \( f_1 \) and the frequency of the last tuning fork as \( f_{56} \).

### Step 1: Define the frequencies
Let the frequency of the first tuning fork be \( f_1 \). Each subsequent tuning fork produces 4 beats per second with the previous one. Therefore, the frequency of the second tuning fork \( f_2 \) can be expressed as:
\[ f_2 = f_1 + 4 \]
Similarly, the frequency of the third tuning fork \( f_3 \) will be:
\[ f_3 = f_2 + 4 = f_1 + 4 + 4 = f_1 + 8 \]
Continuing this pattern, we can express the frequency of the \( n^{th} \) tuning fork as:
\[ f_n = f_1 + 4(n-1) \]

### Step 2: Find the frequency of the last fork
For the 56th tuning fork, we have:
\[ f_{56} = f_1 + 4(56-1) = f_1 + 4 \times 55 = f_1 + 220 \]

### Step 3: Relate the first and last fork frequencies
According to the problem, the frequency of the last fork is three times that of the first fork:
\[ f_{56} = 3f_1 \]

### Step 4: Set up the equation
Now, we can set up the equation using the expressions we derived:
\[ f_1 + 220 = 3f_1 \]

### Step 5: Solve for \( f_1 \)
Rearranging the equation gives:
\[ 220 = 3f_1 - f_1 \]
\[ 220 = 2f_1 \]
\[ f_1 = \frac{220}{2} = 110 \]

### Conclusion
Thus, the frequency of the first tuning fork \( f_1 \) is \( 110 \, \text{Hz} \).

### Final Answer
The frequency of the first fork is \( 110 \, \text{Hz} \).

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