A guitar string is vibrating in its fundamental mode, with nodes at each end. The length of the segment of the string that is free to vibrate is 0.386 m. The maximum transverse acceleration of a point at the middle of the segment is 8.40 × 10 3 m/s 2 and the maximum transverse velocity is 3.80 m/s.
Text Solution
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The displacement of the string at any point is y(x, t) = (A SW sin kx)sin ωt. For the fundamental mode λ = 2L, so at the midpoint of the string
sin kx = sin(2 𝜋 /λ)(L/2) = 1, and
y = A SW sin ωt. Taking derivatives gives v y = 
= ωA SW cos ωt, with maximum value
v y max = ωA SW , and a y = 
= – ω 2 A SW sin ωt, with maximum value a y max = ω 2 A SW .
Dividing these gives
ω = a y max /v y max = (8.40 × 10 3 m/s 2 )/(3.80 m/s) = 2.21 × 10 3 rad/s, and then
A SW = v y max /ω = (3.80 m/s)/(2.21 × 10 3 rad/s) = 1.72 × 10 –3 m.
v = λf = (2L)( ω/2 𝜋 ) = Lω/ 𝜋 = (0.386 m)(2.21 × 10 3 rad/s)/ 𝜋 = 272 m/s.
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