A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 192 m/s and a frequency of 240 Hz. The amplitude of the standing wave at an antinode is 0.400 cm
Text Solution
Verified by Experts(i) (0; (ii) (0; (iii) (0; (i) 0, 0; (ii) 6; (iii) 4
λ = v/f = (192.0 m/s)/(240.0 Hz) = 0.800 m, and the wave amplitude is A SW = 0.400 cm. The amplitude of the motion at the given points is (i) (0.400 cm)sin ( 𝜋 ) = 0 (a node), (ii) (0.400 cm) sin( 𝜋 /2) = 0.004 cm (an antinode) and (iii) (0.400 cm) sin( 𝜋 /4) = 0.283 cm.
The time is half of the period, or 1/(2f) = 2.08 × 10 –3 s.
In each case, the maximum velocity is the amplitude multiplied by ω = 2 𝜋 f and the maximum acceleration is the amplitude multiplied by ω 2 = 4 𝜋 2 f 2 , or (i) 0, 0; (ii) 6.03 m/s, 9.10 × 10 3 m/s 2 ; (iii) 4.27 m/s, 6.43 × 10 3 m/s 2 .
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