Question

A transverse wave is represented by $y=Asin(\omega t-kx)$ . 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
$y=Asin(\omega t-kx)$
$\text{Wave velociy,}\text{v}=\frac{\omega }{k}\dots \left(\text{i}\right)$
$\text{Particle velocity,}{\left(\text{v}\right)}_p=\frac{dy}{dt}=A\omega \text{cos}\left(\omega t-kx\right)$
Maximum particle velocity, ${\left(v_{\text{p}}\right)}_{\text{max}}=\text{A}\omega$ ...(ii)
According to the given question
$v={\left(v_{\text{p}}\right)}_{\text{max}}$
$\frac{\omega }{k}=A\omega \left(\text{Using (i) and (ii)}\right)$
$\frac{1}{k}=A$
$\text{or}\frac{\lambda }{2\pi }=A\left(\because k=\frac{2\pi }{\lambda }\right)$
$\lambda =2\pi A$