Answer:
a) λ = 1.12 m
b) f = 5.41 Hz
c) v = 154.54 m/s
d) A = 0.22m
e)
[tex]v_D_{max}=7.48\frac{m}{s}\\\\v_D_{min}=-7.48\frac{m}{s}\\\\[/tex]
Explanation:
You have the following equation for a wave traveling on a cord:
[tex]D=0.22sin(5.6x+34t)[/tex] (1)
The general expression for a wave is given by:
[tex]D=Asin(kx-\omega t)[/tex] (2)
By comparing the equation (1) and (2) you have:
A: amplitude of the wave = 0.22m
k: wave number = 5.6 m^-1
w: angular velocity = 34 rad/s
a) The wavelength is given by substitution in the following expression:
[tex]\lambda=\frac{2\pi}{k}=\frac{2\pi}{5.6m^{-1}}=1.12m[/tex]
b) The frequency is:
[tex]f=\frac{\omega}{2\pi}=\frac{34s^{-1}}{2\pi}=5.41Hz[/tex]
c) The velocity of the wave is:
[tex]v=\frac{\omega}{k}=\frac{34s^{-1}}{0.22m^{-1}}=154.54\frac{m}{s}[/tex]
d) The amplitude is 0.22m
e) To calculate the maximum and minimum speed of the particles you obtain the derivative of the equation of the wave, in time:
[tex]v_D=\frac{dD}{dt}=(0.22)(34)cos(5.6x+34t)\\\\v_D=7.48cos(5.6x+34t)[/tex]
cos function has a minimum value -1 and maximum +1. Then, you obtain for maximum and minimum velocity:
[tex]v_D_{max}=7.48\frac{m}{s}\\\\v_D_{min}=-7.48\frac{m}{s}\\\\[/tex]
