Question

A transverse wave on a cable is described by the function y(x,t)=2.3cos(4.7x+12t−π/2)y(x,t)=2.3cos(4.7x+12t-π/2), where distance is measured in meters and time in seconds. If the linear mass density of the cable is 0.42 kg/m, what is the average power transmitted by the cable?

Answer 1

P=408.01W

Step-by-step explanation:

We have the wave function:

`y(x,t)=2.3cos(4.7x+12t-(\pi)/(2))` (1)

The average power transmitted by the wave is given by:

`P=(E)/(T)\\\\E=(1)/(2)\mu \omega^2A^2\lambda` (2)

where E is the energy of the wave, mu is the linear mass density, w is the angular frequency, lambda is the wavelength and A is the amplitude.

The general form of a wave equation can be expressed as:

`y(x,t)=Acos(kx-\omega t+\phi)` (3)

by comparing with the equation (1) we obtain that:

`A=2.3\\k=4.7\\\omega=12\\\phi=-(\pi)/(2)\\\lambda=(2\pi)/(k)=0.425\pi`

`T=(2\pi)/(\omega)=0.523s`

Finally, by replacing in (2) we obtain:

`E=(1)/(2)(0.42(kg)/(m))(12s^(-1))^2(2.3m)^2(0.425\pi m)=213.58J\\\\P=(213.58J)/(0.523s)=408.01W`

hope this helps!!