Question #150778
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]
1
Expert's answer
2020-12-21T04:30:25-0500

The wave function can be written as follows:


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


or,


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


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

1)

a) A=2.5 m.A=2.5\ m.

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

c) f=ω2π=180 rads2π=28.65 Hz.f = \dfrac{\omega}{2\pi}=\dfrac{180\ \dfrac{rad}{s}}{2\pi}=28.65\ Hz.

d) λ=2πk=2π20 radm=0.31 m.\lambda = \dfrac{2\pi}{k}=\dfrac{2\pi}{20\ \dfrac{rad}{m}}=0.31\ m.

e) T=1f=128.65 Hz=0.035 s.T=\dfrac{1}{f}=\dfrac{1}{28.65\ Hz}=0.035\ s.

f) v=fλ=28.65 Hz0.31 m=8.88 ms.v=f\lambda=28.65\ Hz\cdot 0.31\ m=8.88\ \dfrac{m}{s}.

d) y(x,t)=(2.5 m)sin(20 radm0.5 m+180 rads2 s)=1.62 m.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.A=0.35\ m.

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

c) f=ω2π=120 rads2π=19.1 Hz.f = \dfrac{\omega}{2\pi}=\dfrac{120\ \dfrac{rad}{s}}{2\pi}=19.1\ Hz.

d) λ=2πk=2π910 radm=0.007 m.\lambda = \dfrac{2\pi}{k}=\dfrac{2\pi}{910\ \dfrac{rad}{m}}=0.007\ m.

e) T=1f=119.1 Hz=0.052 s.T=\dfrac{1}{f}=\dfrac{1}{19.1\ Hz}=0.052\ s.

f) v=fλ=19.1 Hz0.007 m=0.13 ms.v=f\lambda=19.1\ Hz\cdot 0.007\ m=0.13\ \dfrac{m}{s}.

d) y(x,t)=(0.35 m)cos(910 radm0.5 m120 rads2 s)=0.07 m.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.A=0.6\ m.

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

c) f=ω2π=90 rads2π=14.32 Hz.f = \dfrac{\omega}{2\pi}=\dfrac{90\ \dfrac{rad}{s}}{2\pi}=14.32\ Hz.

d) λ=2πk=2π30 radm=0.21 m.\lambda = \dfrac{2\pi}{k}=\dfrac{2\pi}{30\ \dfrac{rad}{m}}=0.21\ m.

e) T=1f=114.32 Hz=0.07 s.T=\dfrac{1}{f}=\dfrac{1}{14.32\ Hz}=0.07\ s.

f) v=fλ=14.32 Hz0.07 m=1.0 ms.v=f\lambda=14.32\ Hz\cdot 0.07\ m=1.0\ \dfrac{m}{s}.

d) y(x,t)=(0.6 m)sin(30 radm0.5 m+90 rads2 s)=0.15 m.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.


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