Question

The figure below shows the wave $y=Asin\left(\right.\omega t-kx\left.\right)$ 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, $y=Asin\left(\right.\omega t-kx\left.\right)$
Slope at any point is given as $\frac{\partial y}{\partial x}|_{t}$
$\Rightarrow $ Slope $=-Akcos\left(\right.\omega t-kx\left.\right)...\left(\right.i\left.\right)$ .
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 $\left(\right.i\left.\right)$ , we can say that
slope $=-Akcos\left[\omega \left(\right. 0 \left.\right) - \frac{2 \pi }{\lambda } \times \frac{\lambda }{2}\right]$ Where, $k=\frac{2 \pi }{\lambda }$
$\Rightarrow $ slope $=-Akcos\pi $
$\Rightarrow $ slope $=Ak$ .