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 find the frequency of the note emitted by the taut string, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The original frequency of the violin string is \( f_0 = 205 \, \text{Hz} \).
- When the string is made slightly more taut, it produces 6 beats in 2 seconds with a tuning fork of frequency 205 Hz.

2. **Calculate the Frequency of Beats**:
- The number of beats produced in 2 seconds is given as 6.
- Therefore, the number of beats per second (beat frequency) is:
\[ \text{Beat frequency} = \frac{6 \, \text{beats}}{2 \, \text{seconds}} = 3 \, \text{Hz} \]

3. **Determine the Possible Frequencies**:
- The beat frequency is the absolute difference between the frequency of the taut string (\( f_t \)) and the frequency of the tuning fork (\( f_0 \)):
\[ |f_t - f_0| = 3 \, \text{Hz} \]
- This gives us two possible equations:
\[ f_t - 205 = 3 \quad \text{or} \quad 205 - f_t = 3 \]

4. **Solve the Equations**:
- From \( f_t - 205 = 3 \):
\[ f_t = 205 + 3 = 208 \, \text{Hz} \]
- From \( 205 - f_t = 3 \):
\[ f_t = 205 - 3 = 202 \, \text{Hz} \]

5. **Determine the Correct Frequency**:
- Since the tension of the string was increased, the frequency must also increase. Therefore, the correct frequency of the taut string is:
\[ f_t = 208 \, \text{Hz} \]

### Final Answer:
The frequency of the note emitted by the taut string is **208 Hz**.

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