Answer with Explanation:
We are given that
Tension=T=233 N
The displacement of the rope is given by
[tex]y=(0.320 m)sin(\frac{\pi x}{3})sin(10\pi)t[/tex]
a.By comparing with
[tex]y=Asin(kx)sin(\omega t)[/tex]
We get
A=0.32
k=[tex]\frac{\pi}{3}[/tex]
[tex]\omega=10\pi[/tex]
[tex]k=\frac{2\pi}{\lambda}[/tex]
[tex]\frac{\pi}{3}=\frac{2\pi}{\lambda}[/tex]
[tex]\lambda=3\times 2=6m[/tex]
n=2
[tex]n\lambda=2L[/tex]
[tex]2\times 6=2L[/tex]
[tex]L=6[/tex]m
b.[tex]\omega=2\pi f[/tex]
[tex]2\pi f=10\pi[/tex]
[tex]f=\frac{10}{2}=5Hz[/tex]
Speed,[tex]v=f\lambda=5\times 6=30m/s[/tex]
c.Let
Mass of the rope=m
[tex]\mu=\frac{m}{L}=\frac{m}{6}[/tex]
[tex]v^2=\frac{T}{\mu}[/tex]
[tex](30)^2=\frac{233}{\frac{m}{6}}[/tex]
[tex]900\times \frac{m}{6}=233[/tex]
[tex]m=\frac{233\times 6}{900}=1.553 kg[/tex]
