Two identical wires are streched by the same tension of `100 N` and each emits a note of frequency `200 H_(Z)` . If the tension in one wire is increased by `1 N` , then the beat frequency is

Options

A. `2 H_(Z)`
B. `(1)/(2) H_(Z)`
C. `1 H_(Z)`
D. None of these

Explanation

To solve the problem step-by-step, we need to determine the new frequency of the first wire after the tension is increased and then calculate the beat frequency between the two wires.

### Step 1: Understand the relationship between tension and frequency
The frequency of a vibrating wire is related to the tension in the wire by the formula:
\[ f \propto \sqrt{T} \]
Where \( f \) is the frequency and \( T \) is the tension.

### Step 2: Calculate the frequency of the first wire after increasing tension
Initially, both wires have a tension of \( T_1 = 100 \, N \) and emit a frequency of \( f_2 = 200 \, Hz \). When the tension in the first wire is increased by \( 1 \, N \), the new tension becomes:
\[ T_1' = 100 \, N + 1 \, N = 101 \, N \]

Using the relationship between frequency and tension, we can express the new frequency \( f_1' \) of the first wire as:
\[ \frac{f_1'}{f_2} = \sqrt{\frac{T_1'}{T_2}} \]
Where \( T_2 = 100 \, N \).

Substituting the values:
\[ \frac{f_1'}{200} = \sqrt{\frac{101}{100}} \]
\[ f_1' = 200 \times \sqrt{\frac{101}{100}} \]

### Step 3: Calculate \( f_1' \)
Now we calculate \( f_1' \):
\[ f_1' = 200 \times \sqrt{1.01} \]

Using the approximation \( \sqrt{1.01} \approx 1.005 \):
\[ f_1' \approx 200 \times 1.005 = 201 \, Hz \]

### Step 4: Calculate the beat frequency
The beat frequency \( f_b \) is given by the absolute difference between the two frequencies:
\[ f_b = |f_1' - f_2| \]
Substituting the values:
\[ f_b = |201 \, Hz - 200 \, Hz| = 1 \, Hz \]

### Final Answer
The beat frequency is \( 1 \, Hz \).

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