A closed organ pipe is sounded near a guitar, causing one of the string to vibrate with large amplitude . We vary the tension of the string until we find the maximum amplitude. The string is `80%` as long as the closed pipe. If both the pipe and the string vibrate at their fundamental frequency, calculate the ratio of the wave speed on the string to the speed of sound in air.

Explanation

To solve the problem, we need to find the ratio of the wave speed on the string (Vt) to the speed of sound in air (Vs) given that both the closed organ pipe and the string vibrate at their fundamental frequencies.

### Step-by-Step Solution:

1. **Define the Lengths**:
- Let the length of the closed organ pipe be \( L \).
- The length of the string is given as \( 80\% \) of the length of the pipe, so:
\[ l = 0.8L \]

2. **Fundamental Frequency of the String**:
- The fundamental frequency \( f \) of the string can be expressed as:
\[ f = \frac{V_t}{2l} \]
- Substituting \( l = 0.8L \):
\[ f = \frac{V_t}{2 \times 0.8L} = \frac{V_t}{1.6L} \]

3. **Fundamental Frequency of the Closed Pipe**:
- The fundamental frequency \( f' \) of a closed pipe is given by:
\[ f' = \frac{V_s}{4L} \]

4. **Equating the Frequencies**:
- Since both the string and the pipe vibrate at the same fundamental frequency, we can set \( f = f' \):
\[ \frac{V_t}{1.6L} = \frac{V_s}{4L} \]

5. **Canceling the Length \( L \)**:
- We can cancel \( L \) from both sides:
\[ \frac{V_t}{1.6} = \frac{V_s}{4} \]

6. **Rearranging to Find the Ratio**:
- Rearranging gives us:
\[ \frac{V_t}{V_s} = \frac{1.6}{4} \]

7. **Calculating the Ratio**:
- Simplifying the right side:
\[ \frac{V_t}{V_s} = \frac{1.6}{4} = \frac{16}{40} = \frac{2}{5} \]

### Final Answer:
The ratio of the wave speed on the string to the speed of sound in air is:
\[ \frac{V_t}{V_s} = \frac{2}{5} \]