64 tuning forks are arranged such that each fork produces 4 beats per second with next one. If the frequency of the last fork is octave of the first, the frequency of 16th fork is

Options

A. 316 Hz
B. 322 Hz
C. 312 Hz
D. 314 Hz

Answer

To solve the problem step-by-step, we can follow these steps:

### Step 1: Understand the Problem
We have 64 tuning forks arranged such that each fork produces 4 beats per second with the next one. The frequency of the last fork (64th fork) is an octave of the first fork (1st fork). We need to find the frequency of the 16th fork.

### Step 2: Define Variables
Let:
- \( f_1 \) = frequency of the first tuning fork
- \( f_{64} \) = frequency of the last tuning fork
- The difference in frequency between two adjacent forks = 4 Hz (since they produce 4 beats per second).

### Step 3: Establish Relationships
From the problem, we know:
1. The frequency of the last fork is twice the frequency of the first fork:
\[ f_{64} = 2f_1 \]
2. The frequency of the nth fork can be expressed as:
\[ f_n = f_1 + (n - 1) \cdot 4 \]
where \( n \) is the number of the fork.

### Step 4: Calculate Frequency of the 64th Fork
Using the formula for the frequency of the nth fork, we can find \( f_{64} \):
\[ f_{64} = f_1 + (64 - 1) \cdot 4 \]
\[ f_{64} = f_1 + 63 \cdot 4 \]
\[ f_{64} = f_1 + 252 \]

### Step 5: Set Up the Equation
Now we have two expressions for \( f_{64} \):
1. \( f_{64} = 2f_1 \)
2. \( f_{64} = f_1 + 252 \)

Setting these equal to each other:
\[ 2f_1 = f_1 + 252 \]

### Step 6: Solve for \( f_1 \)
Subtract \( f_1 \) from both sides:
\[ 2f_1 - f_1 = 252 \]
\[ f_1 = 252 \text{ Hz} \]

### Step 7: Calculate Frequency of the 16th Fork
Now we can find the frequency of the 16th fork using the formula:
\[ f_{16} = f_1 + (16 - 1) \cdot 4 \]
\[ f_{16} = 252 + 15 \cdot 4 \]
\[ f_{16} = 252 + 60 \]
\[ f_{16} = 312 \text{ Hz} \]

### Conclusion
The frequency of the 16th tuning fork is **312 Hz**.

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