A certain string on a piano is tuned to produce musical note of frequency f =261.63 Hz by carefully adjusting the tension in the string. For a fixed wavelength, what is the frequency when this tension is doubled?

Options

A. 130.08 Hz
B. 185.00 Hz
C. 370.00 Hz
D. 446.63 Hz

Answer

To solve the problem, we need to understand how the frequency of a vibrating string is related to the tension in the string. The frequency \( f \) of a vibrating string is given by the formula:

\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

where:
- \( f \) is the frequency,
- \( L \) is the length of the string,
- \( T \) is the tension in the string,
- \( \mu \) is the linear mass density of the string.

### Step 1: Understand the relationship between tension and frequency
From the formula, we can see that the frequency is proportional to the square root of the tension:

\[ f \propto \sqrt{T} \]

This means that if we double the tension, the frequency will change according to the square root of the tension.

### Step 2: Calculate the new frequency when tension is doubled
Let the initial tension be \( T_1 \) and the initial frequency be \( f_1 = 261.63 \, \text{Hz} \). When the tension is doubled, the new tension \( T_2 \) is:

\[ T_2 = 2T_1 \]

Using the proportionality, the new frequency \( f_2 \) can be expressed as:

\[ f_2 = f_1 \sqrt{\frac{T_2}{T_1}} = f_1 \sqrt{2} \]

### Step 3: Substitute the values
Now substituting the value of \( f_1 \):

\[ f_2 = 261.63 \, \text{Hz} \times \sqrt{2} \]

Calculating \( \sqrt{2} \):

\[ \sqrt{2} \approx 1.414 \]

Now substituting this back into the equation:

\[ f_2 = 261.63 \, \text{Hz} \times 1.414 \approx 369.00 \, \text{Hz} \]

### Conclusion
Thus, when the tension in the string is doubled, the new frequency is approximately \( 369 \, \text{Hz} \).

### Final Answer
The frequency when the tension is doubled is \( 369 \, \text{Hz} \).

---