Question

Figure below shows the wave
$y=A\sin (\omega t-kx)$ at any instant travelling in the $+vex$-direction. What is the slope of the curve at $B$?

• A. $kA$
• B. $\omega /A$
• C. $k/A$
• D. $\omega A$

Answer 1

The correct option is A $kA$

Given: $y=A\sin (\omega t-kx)$
As we know,
Particle velocity $\left(v_p\right)=-v\times$ Slope of the graph at that point
The particle velocity is maximum at $B$ and is given by
$\frac{dy}{dt}=(v_p{)}_{max}=\omega A\dots (i)$
Also, wave velocity is given by,
$\frac{dx}{dt}=v=\frac{\omega }{k}\dots \left(ii\right)$
Therefore, from equation $\left(i\right)$and $\left(ii\right)$ we get,
Slope $\frac{dy}{dt}=\frac{\left(vp\right)max}{v}=kA$

Final answer: $\left(c\right)$