A transverse sinusoidal wave is generated at one end of a long horizontal string by a bar that moves the end up and down through a distance by `2.0 cm`. the motion of bar is continuous and is repeated regularly `125` times per second. If klthe distance between adjacent wave crests is observed to be `15.6 cm` and the wave is moving along positive `x-`direction, and at `t=0` the element of the string at `x=0` is at mean position `y =0` and is moving downward, the equation of the wave is best described by

Options

A. `y=(1 cm) sin [(40.3 rad//m) x-(786 rad//s)t]`
B. `y=(2 cm) sin [(40.3 rad//m) x-(786 rad//s)t]`
C. `y=(1 cm) cos [(40.3 rad//m) x-(786 rad//s)t]`
D. `y=(2 cm) cos[(40.3 rad//m) x-(786 rad//s)t]`

Answer

a

Explanation

Amplitude of wave,
`A=(2.0cm)/(2)=1 cm`
frequency of wave, `f=125 Hz`
Wavelength of wave, `lambda=15.6 cm=0.156 m`
Let equation of wave be, `y=A sin (kx-omega t+phi)` where `k=2pi//lambda=40.3 rad//m and omega=2pif=786 rad//s`
Using initial conditions,
`y(0,0)=0=A sin phi`
and `(dely)/(delt)=Aomega cos philt0`
we get, `phi=0`
So, the equation of wave is
`y=(1 cm) sin[(40.3 rad//m)x-(786 rad//s)t]`