25 tunning forks are arranged in series in the order of decreasing frequency. Any two successive forks produce 3 beats/sec. If the frequency of the first turning fork is the octave of the last fork, then the frequency of the 21st fork is

Options

A. 72 Hz
B. 288 Hz
C. 84 Hz
D. 87 Hz

Answer

To solve the problem, we need to find the frequency of the 21st tuning fork given the conditions of the problem. Let's break it down step by step.

### Step 1: Define the frequency of the last tuning fork.
Let the frequency of the last tuning fork (the 25th fork) be \( n \) Hz.

### Step 2: Determine the frequency of the first tuning fork.
Since the first tuning fork is the octave of the last fork, its frequency will be:
\[ f_1 = 2n \]

### Step 3: Establish the frequency of the successive tuning forks.
We know that any two successive forks produce 3 beats per second. This means that the frequency difference between any two successive forks is 3 Hz. Therefore, we can express the frequencies of the tuning forks as follows:
- Frequency of the 1st fork: \( f_1 = 2n \)
- Frequency of the 2nd fork: \( f_2 = f_1 - 3 = 2n - 3 \)
- Frequency of the 3rd fork: \( f_3 = f_2 - 3 = 2n - 6 \)
- Continuing this pattern, the frequency of the \( k \)-th fork can be expressed as:
\[ f_k = 2n - 3(k - 1) \]

### Step 4: Find the frequency of the 25th tuning fork.
For the 25th fork:
\[ f_{25} = 2n - 3(25 - 1) = 2n - 72 \]
Since we defined \( f_{25} = n \), we can set up the equation:
\[ n = 2n - 72 \]

### Step 5: Solve for \( n \).
Rearranging the equation gives:
\[ 72 = 2n - n \]
\[ n = 72 \text{ Hz} \]

### Step 6: Find the frequency of the 21st tuning fork.
Now we can find the frequency of the 21st fork using the formula derived earlier:
\[ f_{21} = 2n - 3(21 - 1) = 2n - 60 \]
Substituting \( n = 72 \):
\[ f_{21} = 2(72) - 60 = 144 - 60 = 84 \text{ Hz} \]

### Final Answer:
The frequency of the 21st tuning fork is \( 84 \) Hz.

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