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

The correct option is C $2\pi A$

Wave velocity$=v=\frac{\lambda }{T}=\frac{\omega \lambda }{2\pi }$

Maximum particle velocity$={\left(v_{max}\right)}_p=A\omega$

Given, $v={\left(v_{max}\right)}_p$

$⟹\frac{\omega \lambda }{2\pi }=A\omega ⟹\lambda =2\pi A$