Answer with Explanation:
We are given that a transverse travelling wave on a cord is represented by
[tex]D=0.51sin(6.1x+76t)[/tex]
Where D and x in meters
t(in seconds)
a.General equation of transverse wave
[tex]y=Asin(kx+\omega t)[/tex]
By comparing we get
A=0.51
[tex]k=6.1[/tex]
[tex]k=\frac{2\pi}{\lambda}[/tex]
Wavelength,[tex]\lambda=\frac{2\pi}{k}=\frac{2\pi}{6.1}=1.03[/tex]
b.[tex]\omega=76[/tex]
Frequency,[tex]f=\frac{\omega}{2\pi}=\frac{76}{2\pi}=12.096Hz[/tex]
c.Velocity,[tex]v=f\lambda=12.096\times 1.03=12.46m/s[/tex]
Direction:Towards negative x- axis
d.Amplitude,A=0.51 m
e.Maximum speed,[tex]v_{max}=A\omega=0.51\times 76=38.76 m/s[/tex]
Minimum speed,[tex]v_{min}=0[/tex]
