Two identical piano strings of length 0.750 m are each, tuned exactly to 500 Hz. The tension in one of the strings is then increased by `21%`. If they are now struck, what is the beat frequency between the fundamental mode of the two strings?

Options

A. 45
B. 50
C. 60
D. 40

Answer

To solve the problem, we need to find the beat frequency between the two piano strings after one of them has its tension increased by 21%. Let's break this down step by step.

### Step 1: Understand the relationship between tension and frequency
The fundamental frequency (f) of a string is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \(L\) is the length of the string,
- \(T\) is the tension in the string,
- \(\mu\) is the linear mass density of the string.

Since both strings are identical, \(\mu\) and \(L\) will be the same for both strings.

### Step 2: Calculate the new tension
Let the original tension of the first string be \(T_1\). After increasing the tension by 21%, the new tension \(T_2\) becomes:
\[ T_2 = T_1 + 0.21 T_1 = 1.21 T_1 \]

### Step 3: Calculate the frequencies of both strings
The frequency of the first string (original tension) is given as:
\[ f_1 = 500 \text{ Hz} \]

For the second string with increased tension:
\[ f_2 = \frac{1}{2L} \sqrt{\frac{T_2}{\mu}} = \frac{1}{2L} \sqrt{\frac{1.21 T_1}{\mu}} = \sqrt{1.21} \cdot \frac{1}{2L} \sqrt{\frac{T_1}{\mu}} = \sqrt{1.21} \cdot f_1 \]

Calculating \(\sqrt{1.21}\):
\[ \sqrt{1.21} = 1.1 \]
Thus,
\[ f_2 = 1.1 \cdot 500 \text{ Hz} = 550 \text{ Hz} \]

### Step 4: Calculate the beat frequency
The beat frequency \(f_b\) is given by the absolute difference between the two frequencies:
\[ f_b = |f_2 - f_1| = |550 \text{ Hz} - 500 \text{ Hz}| = 50 \text{ Hz} \]

### Final Answer
The beat frequency between the fundamental modes of the two strings is:
\[ \boxed{50 \text{ Hz}} \]