A tuning fork of frequency 280 Hz produces 10 beats per see when sounded with a vibrating sonometer string. When the tension in the string increases slightly, it produces 11 beats per see. The original frequency of the vibrating sonometer string is :

Options

A. 269 Hz
B. 291 Hz
C. 270 Hz
D. 290 Hz

Answer

To find the original frequency of the vibrating sonometer string, we can follow these steps:

### Step 1: Understanding the Beat Frequency
When two sound waves of slightly different frequencies are played together, they produce a phenomenon called beats. The beat frequency is equal to the absolute difference between the two frequencies.

Given:
- Frequency of the tuning fork, \( f_t = 280 \, \text{Hz} \)
- Initial beat frequency, \( b_1 = 10 \, \text{Hz} \)

### Step 2: Setting Up the Equation for the First Case
Let the original frequency of the sonometer string be \( f_1 \). The beat frequency can be expressed as:
\[ |f_1 - f_t| = b_1 \]
This gives us two possible equations:
1. \( f_1 - 280 = 10 \) (if \( f_1 > 280 \))
2. \( 280 - f_1 = 10 \) (if \( f_1 < 280 \))

Solving these equations:
1. From \( f_1 - 280 = 10 \):
\[ f_1 = 290 \, \text{Hz} \]
2. From \( 280 - f_1 = 10 \):
\[ f_1 = 270 \, \text{Hz} \]

### Step 3: Analyzing the Second Case
When the tension in the string increases, the frequency of the sonometer string also increases, leading to a new beat frequency:
- New beat frequency, \( b_2 = 11 \, \text{Hz} \)

Now, we can set up the equation for the second case:
\[ |f_2 - f_t| = b_2 \]
Where \( f_2 \) is the new frequency of the sonometer string after the tension is increased.

### Step 4: Finding the New Frequency
Since the frequency of the string increases, we can analyze the two cases again:
1. If \( f_1 = 290 \, \text{Hz} \):
\[ |f_2 - 280| = 11 \]
This gives:
- \( f_2 - 280 = 11 \) → \( f_2 = 291 \, \text{Hz} \)
- \( 280 - f_2 = 11 \) → \( f_2 = 269 \, \text{Hz} \) (not possible since \( f_2 > f_1 \))

2. If \( f_1 = 270 \, \text{Hz} \):
\[ |f_2 - 280| = 11 \]
This gives:
- \( f_2 - 280 = 11 \) → \( f_2 = 291 \, \text{Hz} \)
- \( 280 - f_2 = 11 \) → \( f_2 = 269 \, \text{Hz} \) (not possible since \( f_2 > f_1 \))

### Step 5: Conclusion
Since the only consistent solution is when \( f_1 = 290 \, \text{Hz} \), we conclude that the original frequency of the vibrating sonometer string is:
\[ \boxed{290 \, \text{Hz}} \]