Answer to Question #150778 in Physics for Anne

Question

Find the (a) amplitude, (b) angular frequency, (c) frequency, (d) wavelength, (e) period, (f) speed of the wave, and the (g) transverse displacement at x = 0.5m and t = 2s on the following wave functions:

1.) y(x,t) = (2.5m)sin[(20 rad/m)x + (180 rad/s)t]

2.) y(x,t) = (0.35m)cos[910 rad/m)x - (120 rad/s)t]

3.) y(x,t) = (0.6m)sin[(30 rad/m)x + (90 rad/s)t]

Answer 1

The wave function can be written as follows:

$$y(x,t)=Asin(kx \pm\omega t),$$

or,

$$y(x,t)=Acos(kx \pm\omega t),$$

here, $A$ is the amplitude of the wave, $k=\dfrac{2\pi}{\lambda}$ is the wave number, $\lambda$ is the wavelength of the wave, $\omega$ is the angular frequency, $y(x,t)$ is the transverse displacement.

1)

a) $A=2.5\ m.$

b) $\omega=180\ \dfrac{rad}{s}.$

c) $f = \dfrac{\omega}{2\pi}=\dfrac{180\ \dfrac{rad}{s}}{2\pi}=28.65\ Hz.$

d) $\lambda = \dfrac{2\pi}{k}=\dfrac{2\pi}{20\ \dfrac{rad}{m}}=0.31\ m.$

e) $T=\dfrac{1}{f}=\dfrac{1}{28.65\ Hz}=0.035\ s.$

f) $v=f\lambda=28.65\ Hz\cdot 0.31\ m=8.88\ \dfrac{m}{s}.$

d) $y(x,t)=(2.5\ m)sin(20\ \dfrac{rad}{m}\cdot 0.5\ m + 180\ \dfrac{rad}{s}\cdot 2\ s)=-1.62\ m.$

2)

a) $A=0.35\ m.$

b) $\omega=120\ \dfrac{rad}{s}.$

c) $f = \dfrac{\omega}{2\pi}=\dfrac{120\ \dfrac{rad}{s}}{2\pi}=19.1\ Hz.$

d) $\lambda = \dfrac{2\pi}{k}=\dfrac{2\pi}{910\ \dfrac{rad}{m}}=0.007\ m.$

e) $T=\dfrac{1}{f}=\dfrac{1}{19.1\ Hz}=0.052\ s.$

f) $v=f\lambda=19.1\ Hz\cdot 0.007\ m=0.13\ \dfrac{m}{s}.$

d) $y(x,t)=(0.35\ m)cos(910\ \dfrac{rad}{m}\cdot 0.5\ m - 120\ \dfrac{rad}{s}\cdot 2\ s)=0.07\ m.$

3)

a) $A=0.6\ m.$

b) $\omega=90\ \dfrac{rad}{s}.$

c) $f = \dfrac{\omega}{2\pi}=\dfrac{90\ \dfrac{rad}{s}}{2\pi}=14.32\ Hz.$

d) $\lambda = \dfrac{2\pi}{k}=\dfrac{2\pi}{30\ \dfrac{rad}{m}}=0.21\ m.$

e) $T=\dfrac{1}{f}=\dfrac{1}{14.32\ Hz}=0.07\ s.$

f) $v=f\lambda=14.32\ Hz\cdot 0.07\ m=1.0\ \dfrac{m}{s}.$

d) $y(x,t)=(0.6\ m)sin(30\ \dfrac{rad}{m}\cdot 0.5\ m + 90\ \dfrac{rad}{s}\cdot 2\ s)=-0.15\ m.$