Figure shows the wave $y = A\sin (\omega t - kx)$. What is the magnitude of slope of the curve at $B$ ?

Figure shows the wave $y = A\sin (\omega t - kx)$. What is the magnitude of slope of the curve at $B$ ? A graph showing a sine wave $y = A\sin(\omega t - kx)$ along Y-X axes passing through origin O and crossing the X-axis at point B.
  1. $\frac{\omega}{A}$
  2. $\frac{k}{A}$
  3. $kA$
  4. $\omega A$

Solution

The particle velocity is maximum at $B$ and is given by $\frac{dy}{dt} = (v_p)_{\text{max}} = \omega A$ Also, wave velocity, $\frac{dx}{dt} = v = \frac{\omega}{k}$ So, slope, $\frac{dy}{dx} = \frac{(v_p)_{\text{max}}}{v} = kA$

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