given equation
y = 0.155 sin (0.525x - 46.5t)
compare with wave equation
y =A sin(kx-wt)
Amplitude A =0.155 m
wave number K = 0.525 m
ang velocity w = 46.5 rad/s
mass per unit lenght mu = 13.2 g/m =0.0132 kg/m
A)
maximum transverse acceleration a =-W^2*A
                                 Â
= - 46.5^2*0.155 = - 335.148 m/s^2
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B)
the maximum transverse force on a 1.00-cm segment of the
string
Fmax= mass *acceleration = lenght *mu *a
                                         Â
= 0.01*0.0132*(-335.148) = - 0.044239 N =
-44.2x10^-3 N
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C)
wavespeed v = w/k = sqrt(T/mu)
46.5/0.525 = sqrt(T/0.0132)
Tension T = 103.55 N
Tension/Fmax = 103.55/0.044239 = 2340.75
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