A standing wave is produced in a vibrating string as shown. If the length of the string is 1.5 m and the frequency of the vibrating motor is 60 Hz, the speed of the wave is

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We’re dealing with a standing wave on a string with both ends fixed. For such a string, the allowed wavelengths are λ = 2L / n, where L is the length of the string and n is a positive integer (the number of half-wavelengths along the length). Here, L = 1.5 m, so λ = 2 × 1.5 / n = 3.0 / n meters.

The wave speed v is related to frequency f and wavelength by v = f × λ. The given frequency is f = 60 Hz, so v = 60 × (3.0 / n) = 180 / n m/s.

Now consider each option:

- Option a: 15 m/s. If v were 15 m/s, then λ = v / f = 15 / 60 = 0.25 m. This would require 0.25 m = 3.0 / n, so n = 12. That could be a valid harmonic in principle, but it does not align with the pattern shown in the figure (which we interpret from the problem as having three half-wavelengths along the string). Since 0.25 m would imply many more loops than likely depicted, this option is inconsistent with the given setup.

- Option b: 60 m/s. If v = 60 m/s, then λ = v / f = 60 / 60 = 1.0 m. This gives 1.0 m = 3.0 / n, so n = 3. A third-harmonic standing wave on a 1.5 m string indeed has three half-wavelengths fitting along the length (three loops), which matches a plausible pictured mode.

- Option c: 40 m/s. Then λ = 40 / 60 ≈ 0.666... m. This would require 0.666... = 3.0 / n, so n ≈ 4.5, which is not an integer. Since standing waves on a string with fixed ends occur only at integer n, this option cannot correspond to a valid mode.

- Option d: 20 m/s. Then λ = 20 / 60 = 0.333... m. This would require 0.333... = 3.0 / n, so n ≈ 9. That is an integer, so in principle a ninth-harmonic could exist, but it would correspond to many tightly spaced loops along the string, which is unlikely for the depicted figure unless the mode is clearly that high. Without evidence of nine half-wavelengths in the picture, this is not the expected answer.

- Option e: 90 m/s. Then λ = 90 / 60 = 1.5 m. This would require 1.5 = 3.0 / n, so n = 2. A second-harmonic (two half-wavelengths, i.e., a single antinode in the middle) would be a different mode than what the figure suggests. If the diagram shows more than a simple two-loop pattern, this would be inconsistent with the pictured mode.

Summarizing the speed possibilities from the harmonic relation v = f × λ and λ = 2L / n gives a set of candidate speeds 180/n m/s for integer n. Among the provided options, 60 m/s corresponds to n = 3, i.e., the third harmonic, which aligns with a standing-wave pattern that typically shows three segments (three half-wavelengths) along the string.

MUF0122 Physics Unit 2 - Semester 2, 2025 4.15 Air columns and strings

A standing wave is produced in a vibrating string as shown. If the length of the string is 1.5 m and the frequency of the vibrating motor is 60 Hz, the speed of the wave is

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