Question

A simple harmonic wave is represented by the relation
$y(x,t)=A\sin 2x(vt-\frac{x}{\lambda })$
If the maximum particle velocity is three times the wave velocity, the wavelength $\lambda$ of the wave is

• A. $\frac{\pi A}{3}$
• B. $\frac{2\pi A}{3}$
• C. $\pi A$
• D. $\frac{\pi A}{2}$

Answer 1

The correct option is B $\frac{2\pi A}{3}$

Maximum particle velocity is,
$(v_P{)}_{\text{max}}=A\omega$
Also $\omega =2\pi f$
$\Rightarrow (v_P{)}_{\text{max}}=A\times 2\pi f$
Now for wave velocity,
$v=\frac{\omega }{k}=\frac{\frac{2\pi }{T}}{\frac{2\pi }{\lambda }}=\frac{\lambda }{T}=\lambda f$
$(\because \frac{1}{T}=f)$
Since maximum particle velocity is thrice of wave velocity,
$\Rightarrow (v_P{)}_{\text{max}}=3v$
$\Rightarrow A\times 2\pi f=3\lambda f$
$\Rightarrow \lambda =\frac{2\pi A}{3}$