A tuning fork vibrating with a sonometer having 20 cm wire produces 5 beats per second. The beat frequency does not change if the length of the wire is changed to 21 cm. The frequency of the tuning fork (in Hertz) must be

Options

A. 200
B. 210
C. 205
D. 215

Explanation

To solve the problem step by step, we will analyze the information given and use the relationships between frequency, length, and beat frequency.

### Step 1: Understand the Beat Frequency
The problem states that a tuning fork vibrating with a sonometer wire of length 20 cm produces 5 beats per second. This means that the frequency of the tuning fork (let's denote it as \( N \)) and the frequency of the sonometer wire (let's denote it as \( f_1 \)) differ by 5 Hz.

Thus, we can write:
\[ |N - f_1| = 5 \]
This implies:
\[ f_1 = N - 5 \quad \text{or} \quad f_1 = N + 5 \]

### Step 2: Analyze the Change in Length
When the length of the wire is changed to 21 cm, the beat frequency remains 5 Hz. This means that the new frequency of the sonometer wire (let's denote it as \( f_2 \)) is related to the tuning fork frequency in the same way:
\[ |N - f_2| = 5 \]
Thus, we can write:
\[ f_2 = N - 5 \quad \text{or} \quad f_2 = N + 5 \]

### Step 3: Relate Frequency and Length
The frequency of a vibrating wire is inversely proportional to its length. The fundamental frequency \( f \) of a wire of length \( L \) can be expressed as:
\[ f = \frac{T}{2L} \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension and \( \mu \) is the mass per unit length. Since \( T \) and \( \mu \) are constants for the same wire, we can say:
\[ f \propto \frac{1}{L} \]

Thus, we can express the frequencies for the two lengths:
\[ f_1 = k \cdot \frac{1}{20} \quad \text{and} \quad f_2 = k \cdot \frac{1}{21} \]
where \( k \) is a constant.

### Step 4: Set Up the Equations
From the relationships we established:
1. For \( L = 20 \) cm:
\[ f_1 = k \cdot \frac{1}{20} \]
2. For \( L = 21 \) cm:
\[ f_2 = k \cdot \frac{1}{21} \]

### Step 5: Substitute into the Beat Frequency Equations
Assuming \( f_1 = N - 5 \) and \( f_2 = N + 5 \):
1. From \( f_1 = N - 5 \):
\[ k \cdot \frac{1}{20} = N - 5 \]
Rearranging gives:
\[ N = k \cdot \frac{1}{20} + 5 \]

2. From \( f_2 = N + 5 \):
\[ k \cdot \frac{1}{21} = N + 5 \]
Rearranging gives:
\[ N = k \cdot \frac{1}{21} - 5 \]

### Step 6: Set the Two Expressions for \( N \) Equal
Now we have two expressions for \( N \):
\[ k \cdot \frac{1}{20} + 5 = k \cdot \frac{1}{21} - 5 \]

### Step 7: Solve for \( k \)
Rearranging the equation:
\[ k \cdot \frac{1}{20} - k \cdot \frac{1}{21} = -10 \]
Finding a common denominator (420):
\[ \frac{21k - 20k}{420} = -10 \]
This simplifies to:
\[ \frac{k}{420} = -10 \implies k = -4200 \]

### Step 8: Substitute \( k \) Back to Find \( N \)
Now substitute \( k \) back into either expression for \( N \):
Using \( N = k \cdot \frac{1}{20} + 5 \):
\[ N = -4200 \cdot \frac{1}{20} + 5 = -210 + 5 = -205 \]

### Step 9: Correct the Sign
Since frequency cannot be negative, we need to take the absolute value:
\[ N = 205 \text{ Hz} \]

### Final Answer
The frequency of the tuning fork is:
\[ \boxed{205 \text{ Hz}} \]