Question

A transverse wave is represented by $y=A$sin $(\omega -kx)$.
For what value of the wavelengths 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)

$2\pi A$

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