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 determine the beat frequency between two piano strings after the tension in one of them is increased by 21%. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the relationship between frequency and tension
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 mass per unit length of the string.

Since both strings are identical, \( L \) and \( \mu \) will remain constant.

### Step 2: Calculate the new tension
The tension in one of the strings is increased by 21%. If the original tension is \( T \), the new tension \( T' \) can be expressed as:
\[ T' = T + 0.21T = 1.21T \]

### Step 3: Determine the new frequency
Using the relationship from Step 1, we can express the new frequency \( f' \) for the string with increased tension:
\[ f' = \frac{1}{2L} \sqrt{\frac{T'}{\mu}} = \frac{1}{2L} \sqrt{\frac{1.21T}{\mu}} \]
Since \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), we can relate \( f' \) to \( f \):
\[ f' = \sqrt{1.21} f \]
Calculating \( \sqrt{1.21} \):
\[ \sqrt{1.21} = 1.1 \]
Thus,
\[ f' = 1.1 f \]

### Step 4: Substitute the original frequency
Given that both strings were initially tuned to 500 Hz:
\[ f' = 1.1 \times 500 \text{ Hz} = 550 \text{ Hz} \]

### Step 5: Calculate the beat frequency
The beat frequency \( f_b \) is given by the absolute difference between the two frequencies:
\[ f_b = |f' - f| = |550 \text{ Hz} - 500 \text{ Hz}| = 50 \text{ Hz} \]

### Final Answer
The beat frequency between the two strings is **50 Hz**.

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