When a vibrating tuning fork is placed on a sound box of a sonometer, 8 beats per second are heard when the length of the sonometer wire is kept at 101cm or 100cm. Then the frequency of the tuning frok is (consider that the tension in the wire is kept constant)

Options

A. 1616 Hz
B. 1608 Hz
C. 1632 Hz
D. 1600 Hz

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript:

### Step 1: Understand the Problem
We have a tuning fork placed on a sonometer wire, and we hear 8 beats per second when the lengths of the wire are 101 cm and 100 cm. We need to find the frequency of the tuning fork.

### Step 2: Define Variables
Let:
- \( N \) = frequency of the tuning fork (in Hz)
- \( N_1 \) = frequency of the sonometer wire when the length is 101 cm
- \( N_2 \) = frequency of the sonometer wire when the length is 100 cm

### Step 3: Relationship Between Frequency and Length
The frequency of a vibrating string is inversely proportional to its length, which means:
\[ N_1 \propto \frac{1}{L_1} \quad \text{and} \quad N_2 \propto \frac{1}{L_2} \]
Thus, we can express the relationship as:
\[ N_1 L_1 = N_2 L_2 \]

### Step 4: Express Frequencies in Terms of Tuning Fork Frequency
Since we have the beat frequency of 8 Hz, we can write:
1. For \( L_1 = 101 \) cm:
\[ N - N_1 = 8 \quad \Rightarrow \quad N_1 = N - 8 \]
2. For \( L_2 = 100 \) cm:
\[ N_2 - N = 8 \quad \Rightarrow \quad N_2 = N + 8 \]

### Step 5: Substitute Frequencies into the Length Relationship
Now, substituting \( N_1 \) and \( N_2 \) into the relationship:
\[ (N - 8) \cdot 101 = (N + 8) \cdot 100 \]

### Step 6: Expand and Rearrange the Equation
Expanding both sides:
\[ 101N - 808 = 100N + 800 \]
Rearranging gives:
\[ 101N - 100N = 800 + 808 \]
\[ N = 1608 \text{ Hz} \]

### Conclusion
The frequency of the tuning fork is \( N = 1608 \) Hz.

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