Question

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

• A. $\pi A/2$
• B. $\pi A$
• C. $2\pi A$
• D. $A$

Answer 1

Answer: (C)

The given equation is, $y=A\sin (\omega t-kx)$
Wave velocity, $v=\frac{\omega }{k}$.....(i)
Particle velocity, $v_p=\frac{dy}{dt}=A\omega \cos (\omega t-kx)$
Maximum particle velocity, ${\left(v_p\right)}_{\max }=A\omega$........(ii)
According to the given question, $v={\left(v_p\right)}_{\max }$
$\frac{\omega }{k}=A\omega$
(Using (i) and (ii))
$\frac{1}{k}=A$ or $\frac{\lambda }{2\pi }=A\left(\because k=\frac{2\pi }{\lambda }\right)$