A set of 31 tuning forks is arranged in series of decreasing frequency. Each fork gives 6 beats/sec. with the preceding one. The first fork is the octave of the last. The frequency of the last tuning fork is

Options

A. 120 Hz
B. 150 Hz
C. 360 Hz
D. 180 Hz

Explanation

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

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have 31 tuning forks arranged in a series of decreasing frequency. Each fork produces 6 beats per second with the preceding one. The first fork is an octave of the last fork.

2. **Defining Frequencies**:
Let the frequency of the last tuning fork (31st fork) be \( f \). Since the first fork is an octave of the last, the frequency of the first fork (1st fork) will be \( 2f \).

3. **Frequency Difference**:
The frequency difference between each consecutive fork is such that each fork gives 6 beats per second with the preceding one. This means that the frequency difference between any two consecutive forks is 6 Hz.

4. **Setting Up the Frequency Equation**:
If the frequency of the 31st fork is \( f \), then the frequency of the 30th fork will be \( f + 6 \), the 29th fork will be \( f + 12 \), and so on, until we reach the 1st fork. The frequency of the 1st fork can be expressed as:
\[ f + 6 \times 30 = f + 180 \]
Therefore, the frequency of the 1st fork is:
\[ 2f = f + 180 \]

5. **Solving the Equation**:
Rearranging the equation gives:
\[ 2f - f = 180 \]
\[ f = 180 \text{ Hz} \]

6. **Conclusion**:
The frequency of the last tuning fork is \( 180 \text{ Hz} \).

### Final Answer:
The frequency of the last tuning fork is **180 Hz**.

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