A stretcded sting resomates with tuning fork of frequency 512 Hz. When length o fthe string is 0.5 m. the length of the string required to vibrate resonantly with a tuning fork of frequeny 256 Hz would be

Options

A. 0.25 m
B. 0. 5 m
C. 1 m
D. 2 m

Answer

To solve the problem, we need to determine the length of a stretched string that resonates with a tuning fork of frequency 256 Hz, given that it resonates with a frequency of 512 Hz at a length of 0.5 m.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Frequency 1 (ν₁) = 512 Hz
- Length 1 (L₁) = 0.5 m
- Frequency 2 (ν₂) = 256 Hz
- Length 2 (L₂) = ?

2. **Understand the Relationship:**
- The frequency of a vibrating string is inversely proportional to its length. This means:
\[ \nu \propto \frac{1}{L} \]
- Therefore, we can write the relationship between the two frequencies and their corresponding lengths as:
\[ \frac{\nu_1}{\nu_2} = \frac{L_2}{L_1} \]

3. **Substitute the Known Values:**
- Substitute the known frequencies into the equation:
\[ \frac{512}{256} = \frac{L_2}{0.5} \]

4. **Simplify the Ratio:**
- Simplifying the left side gives:
\[ 2 = \frac{L_2}{0.5} \]

5. **Solve for Length 2 (L₂):**
- To find L₂, multiply both sides by 0.5:
\[ L_2 = 2 \times 0.5 = 1 \text{ m} \]

6. **Conclusion:**
- The length of the string required to vibrate resonantly with a tuning fork of frequency 256 Hz is **1 meter**.