Question

The figure below shows the wave $y=Asin(\omega t-kx)$ at any instant travelling in the +ve $x$ -direction. What is the magnitude of the slope of the curve at $B$ :

• A. $\frac{\omega }{A}$
• B. $\frac{k}{A}$
• C. $kA$
• D. $\omega A$

Answer 1

Answer: (C)

For a sine wave, $y=Asin(\omega t-kx)$
Slope at any point is given as $\frac{\partial y}{\partial x}{|}_t$
$\Rightarrow$ Slope $=-Akcos(\omega t-kx)...(i)$ .
Now, for point B, $x=\frac{\lambda }{2}$ (as we know that two consecutive particles at mean position have a phase difference of $180^{\circ }$ ).
So, using equation $(i)$ , we can say that
slope $=-Akcos\left[\omega (0)-\frac{2\pi }{\lambda }\times \frac{\lambda }{2}\right]$ Where, $k=\frac{2\pi }{\lambda }$
$\Rightarrow$ slope $=-Akcos\pi$
$\Rightarrow$ slope $=Ak$ .