Two stretched wires of same length, diameter and same material are in unison. The tension in one is increased by 2% and 2 beats per second are heard. What was the frequency of the note produced when they were in unison

Options

A. 100 Hz
B. 200 Hz
C. 300 Hz
D. 400 Hz

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Understand the Problem
We have two wires of the same length, diameter, and material, which means they have the same mass per unit length (μ). Initially, they are vibrating in unison, meaning they have the same frequency (F1).

### Step 2: Increase in Tension
When the tension in one wire (let's call it T1) is increased by 2%, the new tension (T2) becomes:
\[ T2 = T1 + 0.02 \cdot T1 = 1.02 \cdot T1 \]

### Step 3: Frequency Relation
The frequency of a vibrating wire is given by the formula:
\[ F = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
Since the length (L) and mass per unit length (μ) are the same for both wires, the frequency is directly proportional to the square root of the tension:
\[ \frac{F2}{F1} = \sqrt{\frac{T2}{T1}} \]

### Step 4: Substitute Tensions
Substituting T2 in terms of T1:
\[ \frac{F2}{F1} = \sqrt{\frac{1.02 \cdot T1}{T1}} = \sqrt{1.02} \]

### Step 5: Calculate the Frequencies
Using the approximation for small changes:
\[ \sqrt{1.02} \approx 1.01 \]
Thus, we can say:
\[ F2 \approx 1.01 \cdot F1 \]

### Step 6: Beat Frequency
The problem states that the beat frequency (the difference between the two frequencies) is 2 beats per second:
\[ F2 - F1 = 2 \]

### Step 7: Substitute for F2
Substituting F2:
\[ 1.01 \cdot F1 - F1 = 2 \]
This simplifies to:
\[ 0.01 \cdot F1 = 2 \]

### Step 8: Solve for F1
Now, solving for F1:
\[ F1 = \frac{2}{0.01} = 200 \text{ Hz} \]

### Conclusion
The frequency of the note produced when the wires were in unison is:
\[ \boxed{200 \text{ Hz}} \]

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