A particular piano wire has middle C (261.7 Hz) for its second harmonic. The wire's length is 2.7m. What is the speed of the wave in the wire? If the wire has 270 lbs of tension what is its linear density? What is the wavelength of this tone emitted into the air? Suppose two such wires emit this tone from two separate locations. A person walks from one wire directly toward the other, noting each point where there is no perceptible sound. What is the smallest separation of two of these points (how close are the neighboring points)?

To answer these questions, we need to use the formulas related to waves and harmonics. Let's break it down step by step:

1. What is the speed of the wave in the wire?
The speed of a wave can be calculated using the formula: v = f * λ, where v is the wave speed, f is the frequency, and λ is the wavelength. In this case, we are given the frequency (261.7 Hz) and the wire's length (2.7m). Since the second harmonic is being considered, it means the wave has two complete cycles in the length of the wire. Therefore, the wavelength (λ) can be calculated by dividing the length of the wire by the number of cycles (2): λ = 2.7m / 2 = 1.35m. Now, using the formula, v = 261.7 Hz * 1.35m = 353.595 m/s.

2. What is the linear density of the wire?
The linear density (μ) can be calculated using the formula: μ = T / v^2, where T is the tension in the wire and v is the wave speed. In this case, the tension is given as 270 lbs (pounds), but we need to convert it to Newtons (SI unit). 1 lb is approximately equal to 4.448 N. Therefore, the tension in Newtons is 270 lbs * 4.448 N/lb = 1,199.76 N. Now, using the formula, μ = 1,199.76 N / (353.595 m/s)^2 = 9.47 kg/m.

3. What is the wavelength of this tone emitted into the air?
The wavelength of the tone emitted into the air is the same as the wavelength in the wire, which we calculated earlier as 1.35m.

4. What is the smallest separation of two points with no perceptible sound?
When a person walks from one wire toward the other, the points where there is no perceptible sound occur at positions where constructive and destructive interference of the waves cancel each other out. These points are called "nodes." The distance between two consecutive nodes is equal to half the wavelength (λ/2). In this case, the wavelength is 1.35m, so the smallest separation between the points would be 1.35m / 2 = 0.675m.

So, the smallest separation between two points with no perceptible sound is 0.675 meters.