Two identical violin strings, when in true and stretched with same tension , have a fundamental frequency of `440.0 H_(Z)`. One of the string is retuned by adjusting its tension . When this is done, `1.5` beats per second are heard when both strings are plucked simultaneously. (a) What are the possible fundamental frequencies of the retuned string? (b) by what fractional amount was the string tension changed if it was (i) increased (ii) decreased?

Explanation

To solve the problem step by step, we will break it down into parts (a) and (b) as per the question.

### (a) Possible Fundamental Frequencies of the Retuned String

1. **Identify the Given Information:**
- Fundamental frequency of the original strings (f₁) = 440 Hz
- Beats per second heard = 1.5 beats/s

2. **Understanding Beats:**
- The number of beats per second is the absolute difference between the frequencies of the two strings.
- If the retuned string has a frequency f₂, then:
\[ |f₂ - f₁| = 1.5 \text{ Hz} \]

3. **Setting Up the Equations:**
- This gives us two possible equations:
\[ f₂ - 440 = 1.5 \quad \text{(1)} \]
\[ 440 - f₂ = 1.5 \quad \text{(2)} \]

4. **Solving the Equations:**
- From equation (1):
\[ f₂ = 440 + 1.5 = 441.5 \text{ Hz} \]
- From equation (2):
\[ f₂ = 440 - 1.5 = 438.5 \text{ Hz} \]

5. **Conclusion for Part (a):**
- The possible fundamental frequencies of the retuned string are **441.5 Hz** and **438.5 Hz**.

### (b) Fractional Change in String Tension

1. **Using the Frequency-Tension Relationship:**
- The frequency of a vibrating string is related to its tension by the formula:
\[ f \propto \sqrt{T} \]
- This implies:
\[ \frac{f₁}{f₂} = \sqrt{\frac{T₁}{T₂}} \]

2. **Rearranging for Tension:**
- Squaring both sides gives:
\[ \left(\frac{f₁}{f₂}\right)^2 = \frac{T₁}{T₂} \]
- Rearranging gives:
\[ T₂ = T₁ \left(\frac{f₂}{f₁}\right)^2 \]

3. **Calculating for Increased Frequency (441.5 Hz):**
- Let’s denote the original tension as T₁.
- For \( f₂ = 441.5 \) Hz:
\[ T₂ = T₁ \left(\frac{441.5}{440}\right)^2 \]
- Calculating:
\[ T₂ = T₁ \left(1.0034\right)^2 \approx T₁ \cdot 1.0068 \]

4. **Calculating Percentage Change:**
- Percentage change in tension:
\[ \text{Percentage Change} = \frac{T₂ - T₁}{T₁} \times 100 = \left(1.0068 - 1\right) \times 100 \approx 0.68\% \]

5. **Calculating for Decreased Frequency (438.5 Hz):**
- For \( f₂ = 438.5 \) Hz:
\[ T₂ = T₁ \left(\frac{438.5}{440}\right)^2 \]
- Calculating:
\[ T₂ = T₁ \left(0.9957\right)^2 \approx T₁ \cdot 0.9932 \]

6. **Calculating Percentage Change:**
- Percentage change in tension:
\[ \text{Percentage Change} = \frac{T₂ - T₁}{T₁} \times 100 = \left(0.9932 - 1\right) \times 100 \approx -0.68\% \]

### Final Answers:
- (a) The possible fundamental frequencies of the retuned string are **441.5 Hz** and **438.5 Hz**.
- (b) The fractional amount of tension changed:
- (i) Increased: **+0.68%**
- (ii) Decreased: **-0.68%**