A transverse sinusoidal wave is moving along a string in the positive direction

Question

A transverse sinusoidal wave is moving along a string in the positive direction of an x axis with a speed of 80 m/s. At t = 0, the string particle at x = 0 has a transverse displacement of 4.0 cm from its equilibrium position and is not moving. The maximum transverse speed of the string particle at x = 0 is 16 m/s.
(a) What is the frequency of the wave?
(b) What is the wavelength of the wave?

If y(x,t) = ym sin(kx ± t + ) is the form of the wave equation,

What are

(c) ym , (d) k, (e) , (f) , (g) the correct choice of sign in front of ?

Explanation / Answer

y(x, t) = ym*sin(kx ± wt +) dy/dt = ±w*ym*cos(kx ± wt + ) the wave is moving along the positive x direction then the correct phase must be: (kx - wt + ). y(0,0)= 0.04 m = ym*sin()   ym = (0.04/sin) (m) Since: 0 = dy/dt(0,0 ) = -w*ym*cos(), must be ±(/2) Thus, ym = 0.04/sin =± 0.04 (m). We take the + sign. The maximum value of dy/dt(0,t): 16 m/s = w*ym = w*ym w = 16/0.04 = 400 (rad/second) (a) the frequency, f = w/(2*) = 63.66 (Hz) (b) the wave length = 2*/k = 2*/(w/c)          = 2**80/400 = 1.2566 m (c) ym = 0.04 (m) (d) k = w/c = 400/80 = 5.0 (e) w = 400 rad/sec (f) If we take ym as positive, then since y(0,t) = 0.04*sin(-400*t), then at t = 0, the displacement is 0.04*sin(± (/2).it has   = /2 (g) When the phase = kx - wt +     0 = d(phase)/dt   (since phase is constant)     = k*dx/dt - w => dx/dt = w/k ( with correct sign)

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