Answer
(a) positive x direction
(b) $\lambda=\frac{2\pi}{B}$
(c) $f=\frac{C}{2\pi}$
(d) $\frac{\pi}{B}$
Work Step by Step
(a) The vertical displacement of a wave is written as follows:
$y(x,t)=Asin\omega(t-\frac{x}{v})$
We know that $(t-\frac{x}{v})$ shows that the wave propagates in the positive x-direction, otherwise it would be $(t+\frac{x}{v})$.
(b) It is given that the vertical displacement of a wave is
$y(x,t)=Asin(Bx-Ct)$....eq(1)
The vertical displacement of a wave having wavelength $\lambda$ and period $T$ is given as
$y(x,t)=Asin(\frac{2\pi}{\lambda}\pm\frac{2\pi}{T}t)$....eq(2)
Comparing eq(1) and eq(2), we obtain:
$B=\frac{2\pi}{\lambda}$
This can be rearranged as:
$\lambda=\frac{2\pi}{B}$
(c) Comparing eq(1) and eq(2), we obtain:
$C=\frac{2\pi}{T}$
$\implies C=2\pi f$
$\implies f=\frac{C}{2\pi}$
(d) We know that for displacement of the wave at $t=0$
$Asin(Bx-C(0))=0$
$\implies AsinBx=0$
$\implies sinBx=sin \space n\pi$
$\implies Bx=n\pi$
$\implies x=\frac{n\pi}{b}$
for $n=1$
$x=\frac{(1)\pi}{B}$
$x=\frac{\pi}{B}$
Thus, the smallest positive value of x is $\frac{\pi}{B}$.
