A violin string vibrates at 298 Hz when unfingered. At what frequency will it vibrate if it is fingered one sixth of the way down from the end?

3/2 (298)= 441

To solve this problem, we need to understand how the length of a vibrating string affects its frequency. The frequency of a vibrating string is inversely proportional to its length. When a string is divided into smaller segments, the frequency increases.

Here are the steps to solve this problem:

1. Determine the length ratio: The question states that the string is fingered one-sixth of the way down from the end. Since we are given the initial length of the string, which is the full length when unfingered, we can calculate the length ratio as follows:
- The vibrating segment length is one-sixth of the full length.
- The remaining length (unvibrated portion) is five-sixths of the full length.
- Therefore, the length ratio is 5:1 (five-sixths to one-sixth).

2. Understand the relationship between length and frequency: The frequency is inversely proportional to the length. In other words, as the length decreases, the frequency increases.

3. Set up the inverse proportion equation: Let's use "f" to represent the frequency when fingered and "f₀" for the initial frequency when unfingered. The proportion equation is then:
- f/f₀ = length₀/length

4. Plug in the given values: We are given that the initial frequency when unfingered is 298 Hz. Let's represent it as "f₀ = 298 Hz." The length ratio is 5:1.

5. Solve the proportion equation: Substituting the given values into the equation, we get:
- f/298 Hz = 1/5

To isolate "f," we cross-multiply to get:
- f = (1/5) * 298 Hz
- f ≈ 59.6 Hz

Therefore, when the violin string is fingered one-sixth of the way down from the end, it will vibrate at approximately 59.6 Hz.