A set of 10 tuning forks is arranged in series of increasing frequency. If each fork gives 3 beats with the preceding one and the last fork has twice the frequency of the first, then frequency of the first tuning fork is

Options

A. 30 Hz
B. 27 Hz
C. 33 Hz
D. 15 Hz

Answer

To solve the problem, we need to analyze the information given about the tuning forks and their frequencies step by step.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a series of 10 tuning forks arranged in increasing frequency. Each fork produces 3 beats with the preceding one, and the last fork has twice the frequency of the first fork.

2. **Defining Variables**:
Let the frequency of the first tuning fork be \( f \).

3. **Finding Frequencies of the Tuning Forks**:
Since each fork gives 3 beats with the preceding one, the frequency of the second fork will be:
\[ f_2 = f + 3 \]
The frequency of the third fork will be:
\[ f_3 = f + 6 \]
Continuing this pattern, the frequency of the \( n \)-th fork can be expressed as:
\[ f_n = f + 3(n - 1) \]
Therefore, the frequency of the 10th fork (when \( n = 10 \)) will be:
\[ f_{10} = f + 3(10 - 1) = f + 27 \]

4. **Using the Given Condition**:
According to the problem, the frequency of the last fork (10th fork) is twice the frequency of the first fork:
\[ f + 27 = 2f \]

5. **Solving for \( f \)**:
Rearranging the equation:
\[ 27 = 2f - f \]
\[ 27 = f \]

6. **Conclusion**:
Thus, the frequency of the first tuning fork is:
\[ f = 27 \text{ Hz} \]

### Final Answer:
The frequency of the first tuning fork is **27 Hz**.

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