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

Answer

To solve the problem, we need to determine the frequency of the tuning fork based on the information given about the sonometer wire lengths and the beat frequency. Let's break it down step by step.

### Step 1: Understand the relationship between frequency and length
The fundamental frequency \( f \) of a vibrating wire is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( L \) is the length of the wire,
- \( T \) is the tension in the wire,
- \( \mu \) is the mass per unit length of the wire.

### Step 2: Set up the equations for the two lengths
We have two lengths of wire: \( L_1 = 20 \, \text{cm} \) and \( L_2 = 21 \, \text{cm} \). The frequencies corresponding to these lengths can be expressed as:
- For \( L_1 \): \( f_1 = n + 5 \) (since it produces 5 beats per second with the tuning fork)
- For \( L_2 \): \( f_2 = n - 5 \)

### Step 3: Use the relationship between frequency and length
From the relationship of frequencies and lengths, we can write:
\[ f_1 \cdot L_1 = f_2 \cdot L_2 \]
Substituting the values we have:
\[ (n + 5) \cdot 20 = (n - 5) \cdot 21 \]

### Step 4: Expand and simplify the equation
Expanding both sides gives:
\[ 20n + 100 = 21n - 105 \]

### Step 5: Rearranging the equation
Now, let's rearrange the equation to isolate \( n \):
\[ 100 + 105 = 21n - 20n \]
\[ 205 = n \]

### Step 6: Conclusion
Thus, the frequency of the tuning fork \( n \) is:
\[ n = 205 \, \text{Hz} \]

### Final Answer
The frequency of the tuning fork is **205 Hz**.

---