A string of mass 0.1 kgm ⁻¹ has length 0.9 m . It is fixed at both ends and stretched such that it has a tension of 40 N . The string vibrates in three segments with amplitude 0.3 cm . The amplitude (maximum) of the particle velocity is (in m/s)

Step-by-step solution

Maximum particle velocity amplitude is determined from the angular frequency and vibration amplitude: u_ max = A . The wave speed is v = T = 40 0.1 = 20 m/s. For the third harmonic on a fixed string, wavelength is = 2L n = 2 0.9 3 = 0.6 m. Frequency becomes f = v = 20 0.6 = 100 3 Hz, yielding angular frequency = 2 f = 200 3 rad/s. Given vibration amplitude A = 0.3 10⁻² m, then u_ max = A = (0.3 10⁻² ) 200 3 = 60 300 = 5 m/s. This corresponds to option C .