Answer:
The average power of this wave is 5.42 Watts.
Explanation:
The transverse wave on a rope is given by :
[tex]y(x,t)=0.75\ cos(\pi [(0.4)x+250t])[/tex]
The general equation of the transverse wave is :
[tex]y=A\ cos(kx+\omega t)[/tex]
Let the mass per unit length of the rope is 0.0500 kg/m.
The average power of this wave is given by :
[tex]P=\dfrac{1}{2}\mu \omega^2A^2 v[/tex]
Where
[tex]\mu[/tex] is the linear mass density, [tex]\mu=0.05\ kg/m[/tex]
[tex]\omega=250\pi = 785.39\ rad/s[/tex]
A = 0.75 cm = 0.0075 m
Propagation constant,
[tex]k=0.4\pi[/tex]
[tex]\dfrac{2\pi}{\lambda}=0.4\pi[/tex]
[tex]\lambda=5\ cm=0.05\ m[/tex]
Angular velocity,
[tex]2\pi f=250\pi[/tex]
f = 125 Hz
Velocity,
[tex]v=f\lambda[/tex]
[tex]v=125\times 0.05=6.25\ m/s[/tex]
Average power of the wave is :
[tex]P=\dfrac{1}{2}\mu \omega^2A^2 v[/tex]
[tex]P=\dfrac{1}{2}\times 0.05\times (785.39)^2 \times (0.0075)^2\times 6.25[/tex]
P = 5.42 Watts
So, the average power of this wave is 5.42 Watts. Hence, this is the required solution.
