The string of a violin emits a note of 205 Hz when its tension is correct. The string is made slightly more taut and it produces 6 beats in 2 seconds with a tuning fork of frequency 205 Hz. The frequency of the note emitted by the taut string is

Options

A. 211 Hz
B. 199 Hz
C. 208 Hz
D. 202 Hz

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the concept of beats
When two sound waves of slightly different frequencies interfere, they produce a phenomenon known as beats. The number of beats per second is equal to the absolute difference between the two frequencies.

### Step 2: Identify the given information
- The frequency of the tuning fork (f_tuning fork) = 205 Hz
- The number of beats produced in 2 seconds = 6 beats
- Therefore, the number of beats in 1 second (f_beats) = 6 beats / 2 seconds = 3 beats per second.

### Step 3: Relate beats to frequency difference
The frequency difference (Δf) between the two sound sources (the taut string and the tuning fork) is equal to the number of beats per second:
\[ \Delta f = f_{string} - f_{tuning fork} \]
Since the frequency of the string is higher than that of the tuning fork (because the tension is increased), we can write:
\[ f_{string} - 205 \text{ Hz} = 3 \text{ Hz} \]

### Step 4: Solve for the frequency of the taut string
Now we can solve for \(f_{string}\):
\[ f_{string} = 205 \text{ Hz} + 3 \text{ Hz} = 208 \text{ Hz} \]

### Conclusion
The frequency of the note emitted by the taut string is **208 Hz**.

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