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 step by step, we will use the information given about the tuning forks and their frequencies.

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

### Step 2: Define the frequency of the last tuning fork
According to the problem, the last tuning fork has twice the frequency of the first one. Therefore, the frequency of the last tuning fork is:
\[ \text{Frequency of last fork} = 2N \text{ Hz} \]

### Step 3: Understand the beat frequency
The problem states that each fork gives 3 beats with the preceding one. This means that the difference in frequency between any two adjacent tuning forks is 3 Hz.

### Step 4: Set up the frequencies of the tuning forks
Since there are 10 tuning forks arranged in increasing order of frequency, we can express the frequencies of the tuning forks as follows:
- 1st fork: \( N \)
- 2nd fork: \( N + 3 \)
- 3rd fork: \( N + 6 \)
- 4th fork: \( N + 9 \)
- 5th fork: \( N + 12 \)
- 6th fork: \( N + 15 \)
- 7th fork: \( N + 18 \)
- 8th fork: \( N + 21 \)
- 9th fork: \( N + 24 \)
- 10th fork: \( N + 27 \)

### Step 5: Relate the last fork to the first fork
From the previous step, we know that the frequency of the 10th fork is \( N + 27 \) and we also know from Step 2 that it must equal \( 2N \). Therefore, we can set up the equation:
\[ N + 27 = 2N \]

### Step 6: Solve for \( N \)
Rearranging the equation gives:
\[ 27 = 2N - N \]
\[ 27 = N \]

### Conclusion
Thus, the frequency of the first tuning fork is:
\[ \boxed{27 \text{ Hz}} \]

---