## College Physics (4th Edition)

(a) Wave I has the fastest speed of $1.50~cm/s$ (b) Wave II has the longest wavelength with a wavelength of $2.09~cm$ (c) Wave II has the fastest maximum speed of a point in the medium with a speed of $13.5~cm/s$ (d) Wave II is moving in the positive x-direction.
In general: $y(x,t) = A~sin(k~x+\omega~t)$ $I. ~y(x,t) = (1.50~cm)~sin~[~(4.00~cm^{-1})~x+(6.00~s^{-1})~t)~]$ $II. ~y(x,t) = (4.50~cm)~sin~[~(3.00~cm^{-1})~x-(3.00~s^{-1})~t)~]$ (a) $v_1 = \frac{\omega}{k} = \frac{6.00~s^{-1}}{4.00~cm^{-1}} = 1.50~cm/s$ $v_2 = \frac{\omega}{k} = \frac{3.00~s^{-1}}{3.00~cm^{-1}} = 1.00~cm/s$ Wave I has the fastest speed of $1.50~cm/s$ (b) $\lambda_1 = \frac{2\pi}{k} = \frac{2\pi}{4.00~cm^{-1}} = 1.57~cm$ $\lambda_2 = \frac{2\pi}{k} = \frac{2\pi}{3.00~cm^{-1}} = 2.09~cm$ Wave II has the longest wavelength with a wavelength of $2.09~cm$ (c) We can find the maximum speed of wave I: $v_m = A~\omega$ $v_m = (1.50~cm)(6.00~s^{-1})$ $v_m = 9.00~cm/s$ We can find the maximum speed of wave II: $v_m = A~\omega$ $v_m = (4.50~cm)(3.00~s^{-1})$ $v_m = 13.5~cm/s$ Wave II has the fastest maximum speed of a point in the medium with a speed of $13.5~cm/s$ (d) The term $(kx-\omega~t)$ shows that a wave is moving in the positive x-direction. Since wave II has the term $(3.00~cm^{-1})~x-(3.00~s^{-1})~t$, we can see that wave II is moving in the positive x-direction.