A set of 24 tuning forks are so arranged that each gives 6 beats per second with the previous one. If the frequency of the last tuning fork is double that of the first, frequency of the second tuning fork is

Options

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

Explanation

To solve the problem, we need to find the frequency of the second tuning fork given the conditions of the problem. Let's break it down step by step.

### Step 1: Define the frequencies
Let the frequency of the first tuning fork be \( f_1 = f \). According to the problem, each tuning fork gives 6 beats per second with the previous one. Therefore, the frequency of the second tuning fork \( f_2 \) will be:
\[ f_2 = f + 6 \]

### Step 2: Determine the frequency of the third tuning fork
Continuing this pattern, the frequency of the third tuning fork \( f_3 \) will be:
\[ f_3 = f_2 + 6 = (f + 6) + 6 = f + 12 \]

### Step 3: Generalize for the nth tuning fork
We can generalize this for the nth tuning fork. The frequency of the nth tuning fork \( f_n \) can be expressed as:
\[ f_n = f + 6(n - 1) \]
where \( n \) is the number of the tuning fork.

### Step 4: Calculate the frequency of the last tuning fork
Since we have 24 tuning forks, the frequency of the last (24th) tuning fork \( f_{24} \) will be:
\[ f_{24} = f + 6(24 - 1) = f + 6 \times 23 = f + 138 \]

### Step 5: Use the given condition
According to the problem, the frequency of the last tuning fork is double that of the first:
\[ f_{24} = 2f \]
Substituting the expression for \( f_{24} \):
\[ f + 138 = 2f \]

### Step 6: Solve for \( f \)
Rearranging the equation gives:
\[ 138 = 2f - f \]
\[ f = 138 \]

### Step 7: Find the frequency of the second tuning fork
Now that we have \( f \), we can find the frequency of the second tuning fork \( f_2 \):
\[ f_2 = f + 6 = 138 + 6 = 144 \text{ Hz} \]

### Final Answer
The frequency of the second tuning fork is \( 144 \text{ Hz} \).

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