Question

A transverse wave is described by the equation, $y=A\sin 2\pi (nt-\frac{x}{{\lambda }_0})$. The maximum particle velocity is equal to $3$ times the wave velocity, if

• A. ${\lambda }_0=\frac{\pi A}{3}$
• B. ${\lambda }_0=\frac{2\pi A}{3}$
• C. ${\lambda }_0=\pi A$
• D. ${\lambda }_0=3\pi A$

Answer 1

The correct option is B ${\lambda }_0=\frac{2\pi A}{3}$

Given that,
$y=A\sin 2\pi (nt-\frac{x}{{\lambda }_0})$
Here, $\omega =2\pi n$ and $k=\frac{2\pi }{{\lambda }_0}$
Maximum particle velocity will be,
$v_{\text{max}}=\omega A=2\pi nA$
Wave velocity will be,
$v_{\text{wave}}=\frac{\omega }{k}=\frac{2\pi n}{\frac{2\pi }{{\lambda }_0}}=n{\lambda }_o$
According to question,
$v_{\text{max}}=3v_{\text{wave}}$
$\Rightarrow 2\pi nA=3\times n{\lambda }_o$
$\Rightarrow {\lambda }_0=\frac{2\pi A}{3}$