Answer
(a) $y(x,t) = (2.0~cm)~cos~[~(1.57\times 10^{-3}~rad/m)~x+(1.57~rad/s)~t)~]$
(b) $v_m = 3.14~cm/s$
(c) $v = 1000~m/s$
Work Step by Step
(a) In general: $y(x,t) = A~cos(k~x+\omega~t)$
It is given that the amplitude is $A = 2.0~cm$
We can find the wave number:
$k = \frac{2\pi}{\lambda}$
$k = \frac{2\pi}{4000~m}$
$k = 1.57\times 10^{-3}~rad/m$
We can find $\omega$:
$\omega = \frac{2\pi}{T}$
$\omega = \frac{2\pi}{4.0~s}$
$\omega = 1.57~rad/s$
We can write an equation for the wave:
$y(x,t) = A~cos(k~x+\omega~t)$
$y(x,t) = (2.0~cm)~cos~[~(1.57\times 10^{-3}~rad/m)~x+(1.57~rad/s)~t)~]$
(b) We can find the maximum speed of the ground:
$v_m = A~\omega$
$v_m = (2.0~cm)(1.57~rad/s)$
$v_m = 3.14~cm/s$
(c) We can find the wave speed:
$v = \frac{\omega}{k}$
$v = \frac{1.57~rad/s}{1.57\times 10^{-3}~rad/m}$
$v = 1000~m/s$
