Missing data: the wave number
[tex]k=60 cm^{-1}[/tex]
(a) [tex]z = 0.003 sin (6000y-31.4 t)[/tex]
For a transverse wave travelling in the positive y-direction and with vibration along the z-direction, the equation of the wave is
[tex]z = A sin (ky-\omega t)[/tex]
where
A is the amplitude of the wave
k is the wave number
[tex]\omega[/tex] is the angular frequency
t is the time
In this situation:
A = 3.0 mm = 0.003 m is the amplitude
[tex]k = 60 cm^{-1} = 6000 m^{-1}[/tex] is the wave number
[tex]T = 0.20 s[/tex] is the period, so the angular frequency is
[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{0.20}=31.4 rad/s[/tex]
So, the wave equation (in meters) is
[tex]z = 0.003 sin (6000y-31.4 t)[/tex]
(b) 0.094 m/s
For a transverse wave, the transverse speed is equal to the derivative of the displacement of the wave, so in this case:
[tex]v_t = z' = -A \omega cos (ky-\omega t)[/tex]
So the maximum transverse wave occurs when the cosine term is equal to 1, therefore the maximum transverse speed must be
[tex]v_{t}_{max} =\omega A[/tex]
where
[tex]\omega = 31.4 rad/s\\A = 0.003 m[/tex]
Substituting,
[tex]v_{t}_{max}=(31.4)(0.003)=0.094 m/s[/tex]
(c) 5.24 mm/s
The wave speed is given by
[tex]v=f \lambda[/tex]
where
f is the frequency of the wave
[tex]\lambda[/tex] is the wavelength
The frequency can be found from the angular frequency:
[tex]f=\frac{\omega}{2\pi}=\frac{31.4}{2\pi}=5 Hz[/tex]
While the wavelength can be found from the wave number:
[tex]\lambda = \frac{2\pi}{k}=\frac{2\pi}{6000}=1.05\cdot 10^{-3} m[/tex]
Therefore, the wave speed is
[tex]v=(5)(1.05\cdot 10^{-3} )=5.24 \cdot 10^{-3} m/s = 5.24 mm/s[/tex]
