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Question

A long string having a cross-sectional area 0.80mm2 and density, 12.5g/cc is subjected to a tension of 64N along the positive x-axis. One end of this string is attached to a vibrator at x=0 moving in transverse direction at a frequency of 20Hz. At t=0, the source is at a maximum displacement y=1.0cm. Write the equation for the wave.

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Solution

Mass per unit length of the string=μ=ρA=0.8×106×12.5×103=0.01kg/m
Thus speed of the wave=Tμ=640.01=80m/s
Amplitude of the wave=A=1cm
ω=2πν=40πs1
v=ωk
k=40π80=π2m1
Thus the wave equation is y=Acos(ωtkx)
=(1cm)cos[(40πs1)t(π2m1)x]

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