A piano wire A vibrates at a fundamental frequency of 600 Hz. A second identical wire B produces 6 beats per second with it when the tension in A is slightly increased. Find the ratio of the tension in A to the tension in B.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We know that:
- The fundamental frequency of wire A, \( \nu_A = 600 \, \text{Hz} \).
- Wire B produces 6 beats per second with wire A when the tension in A is slightly increased.
- The relationship between the frequencies and tension of the wires is given by the formula for the frequency of a vibrating wire.

### Step 2: Use the Beat Frequency Concept
The beat frequency \( n \) is given by:
\[ |\nu_A - \nu_B| = n \]
Here, \( n = 6 \, \text{Hz} \). Since the frequency of wire A increases when the tension is increased, we can write:
\[ \nu_A - \nu_B = 6 \]
This implies:
\[ \nu_A = \nu_B + 6 \]

### Step 3: Substitute the Known Frequency
Since the initial frequency of wire A is \( 600 \, \text{Hz} \):
\[ 600 = \nu_B + 6 \]
Solving for \( \nu_B \):
\[ \nu_B = 600 - 6 = 594 \, \text{Hz} \]

### Step 4: Relate Frequencies to Tensions
Using the relationship between frequency and tension:
\[ \nu = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
Since both wires are identical, we can assume that \( L \) (length) and \( \mu \) (mass per unit length) are the same for both wires. Thus, the ratio of frequencies is related to the ratio of tensions:
\[ \frac{\nu_A}{\nu_B} = \sqrt{\frac{T_A}{T_B}} \]

### Step 5: Square Both Sides
Squaring both sides gives:
\[ \left(\frac{\nu_A}{\nu_B}\right)^2 = \frac{T_A}{T_B} \]

### Step 6: Substitute the Frequencies
Substituting \( \nu_A = 606 \, \text{Hz} \) (after the increase in tension) and \( \nu_B = 594 \, \text{Hz} \):
\[ \frac{T_A}{T_B} = \left(\frac{606}{594}\right)^2 \]

### Step 7: Calculate the Ratio
Calculating the ratio:
\[ \frac{606}{594} \approx 1.0202 \]
Squaring this gives:
\[ \frac{T_A}{T_B} \approx (1.0202)^2 \approx 1.0408 \]

### Final Answer
Thus, the ratio of the tension in wire A to the tension in wire B is approximately:
\[ \frac{T_A}{T_B} \approx 1.02 \]

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