## College Physics (4th Edition)

(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) $v = 1000~m/s$
(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: $v = \frac{\omega}{k}$ $v = \frac{1.57~rad/s}{1.57\times 10^{-3}~rad/m}$ $v = 1000~m/s$