Question

A transverse wave is represented by $y \, = \, A \, sin\left(\right.\omega t \, - \, kx\left.\right)$ . For what value of the wavelength is the wave velocity equal to the maximum particle velocity?

• A. $\frac{\pi A}{2}$
• B. $\pi A$
• C. $2\pi A$
• D. $A$

Answer 1

Answer: (C)

The given wave equation is
$y=Asin\left(\right.\omega t-kx\left.\right)$
$\text{Wave velociy,}\textit{v}=\frac{\omega }{k}\ldots \left(\text{i}\right)$
$\text{Particle velocity,}\left(\textit{v}\right)_{p }=\frac{d ⁡ y}{d ⁡ t}=A⁡\omega \text{cos}\left(\omega t - k x\right)$
Maximum particle velocity, $\left(v_{\text{p}}\right)_{\text{max}}=\textit{A}\omega $ ...(ii)
According to the given question
$v=\left(v_{\text{p}}\right)_{\text{max}}$
$\frac{\omega }{k} = A \omega \left(\text{Using (i) and (ii)}\right)$
$\frac{1}{k} = A $
$\text{or} \frac{\lambda }{2 \pi } = A \left(\because k = \frac{2 \pi }{\lambda }\right)$
$\lambda =2\pi A$