A 12 Hz transverse wave of amplitude 3.0 cm travels at 4.8 m/s in a light string, then at 3.2 m/s in a heavier string. (a) Find both wavelengths. (b) Find period, angular frequency and maximum particle speed. (c) Explain unchanged frequency, changed wavelength and partial reflection. (d) State reflected-pulse phase change.

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Wavelengths 0.400 m and 0.267 m; period 0.0833 s; angular frequency 75.4 rad/s; maximum particle speed 2.26 m/s; reflected pulse is inverted. From v = fλ, wavelengths are 4.8/12 = 0.400 m and 3.2/12 = 0.267 m. Period T = 1/12 = 0.0833 s and angular frequency ω = 2πf = 75.4 rad/s. A string point has maximum transverse speed ωA = 75.4 × 0.030 = 2.26 m/s, distinct from propagation speed. The source fixes frequency and boundary displacement remains continuous. Lower transmitted speed therefore gives shorter wavelength; impedance mismatch requires reflected and transmitted components. Reflection from the higher-impedance, more fixed-like boundary is inverted, a phase change of π. Energy is shared, not duplicated, between reflected and transmitted waves.

Wave transmission at a string boundary

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