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 Hz
B. 210 Hz
C. 205 Hz
D. 215 Hz

Answer

To solve the problem step by step, we will analyze the situation involving the tuning fork and the sonometer wire.

### Step 1: Understanding the Beat Frequency
When the tuning fork and the sonometer wire vibrate together, they produce beats. The beat frequency is the absolute difference between the frequency of the tuning fork (f1) and the frequency of the sonometer wire (fi). Given that the beat frequency is 5 beats per second, we can write:
\[ |fi - f1| = 5 \]

### Step 2: Setting Up the Frequencies
Assuming that the frequency of the sonometer wire (fi) is greater than the frequency of the tuning fork (f1), we can express this as:
\[ fi - f1 = 5 \]
This is our first equation.

### Step 3: Changing the Length of the Wire
When the length of the wire is increased from 20 cm to 21 cm, the frequency of the sonometer wire will decrease because frequency is inversely proportional to the length of the wire. Therefore, we can denote the new frequency of the sonometer wire as fi'.

### Step 4: Relating the Frequencies
Since the beat frequency remains the same (5 beats per second) with the new length, we have:
\[ fi' - f1 = 5 \]
This is our second equation.

### Step 5: Using the Inverse Proportionality
Since the frequency is inversely proportional to the length, we can write:
\[ \frac{fi}{fi'} = \frac{L'}{L} \]
Where L' = 21 cm and L = 20 cm. Thus:
\[ \frac{fi}{fi'} = \frac{21}{20} \]
From this, we can express fi' in terms of fi:
\[ fi' = \frac{20}{21} fi \]

### Step 6: Substituting into the Beat Frequency Equation
Now we can substitute fi' into the second equation:
\[ \frac{20}{21}fi - f1 = 5 \]
Rearranging gives:
\[ \frac{20}{21}fi = f1 + 5 \]

### Step 7: Combining the Two Equations
Now we have two equations:
1. \( fi - f1 = 5 \)
2. \( \frac{20}{21}fi = f1 + 5 \)

From the first equation, we can express fi as:
\[ fi = f1 + 5 \]

### Step 8: Substituting for fi
Substituting this expression for fi into the second equation:
\[ \frac{20}{21}(f1 + 5) = f1 + 5 \]

### Step 9: Solving for f1
Expanding and simplifying:
\[ \frac{20}{21}f1 + \frac{100}{21} = f1 + 5 \]
Multiplying through by 21 to eliminate the fraction:
\[ 20f1 + 100 = 21f1 + 105 \]
Rearranging gives:
\[ 100 - 105 = 21f1 - 20f1 \]
\[ -5 = f1 \]
Thus:
\[ f1 = 205 \, \text{Hz} \]

### Final Answer
The frequency of the tuning fork is:
\[ f1 = 205 \, \text{Hz} \]

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