When a stretched string is clamped at both ends, its fundamental frequency is 140 Hz. (a)What is the next highest frequency ? Sketch the standing wave pattern. (b) If the tension inthe string is 280 N and its mass is 7.1 g what is its length ? The same string is now clamped atone end and is free at the other. Assuming the same tension. (c) Determine the fundamentalfrequency and sketch the standing wave pattern. (d) Determine the next highest frequencyand sketch its standing wave pattern

(a) The next highest frequency is the second harmonic, which is twice the fundamental frequency. So, the next highest frequency is 2 * 140 Hz = 280 Hz. The standing wave pattern would have two antinodes instead of one.

(b) The speed of the wave on the string can be found using the formula v = sqrt(T/μ), where T is the tension and μ is the linear mass density. The linear mass density can be found by dividing the mass by the length of the string. We can rearrange the formula to solve for the length: L = m / (v^2 * μ).

First, we need to find the speed of the wave. The speed is related to the frequency and the wavelength by the formula v = f * λ. For the fundamental frequency, the wavelength is twice the length of the string, so v = 140 Hz * 2L.

We can substitute this into our formula for the length: L = m / (f^2 * 4L * μ). Solving for L gives L = sqrt(m / (4f^2 * μ)).

Substituting the given values gives L = sqrt(7.1 g / (4 * (140 Hz)^2 * 280 N)) = 0.5 m.

(c) When the string is clamped at one end and free at the other, the fundamental frequency corresponds to a wavelength four times the length of the string. The frequency can be found using the formula f = v / λ = v / (4L) = (140 Hz * 2L) / (4L) = 70 Hz. The standing wave pattern would have one antinode at the free end and a node at the clamped end.

(d) The next highest frequency would be three times the fundamental frequency, or 3 * 70 Hz = 210 Hz. The standing wave pattern would have two antinodes and one node.