Question

A transverse wave is described by the equation $y=y_0\sin 2\pi (ft-\frac{x}{t})$. The maximum velocity of the particle is equal to four times the wave velocity, if

• A. $\lambda =\pi y_0/4$
• B. $\lambda =\pi y_0/2$
• C. $\lambda =\pi y_0$
• D. $\lambda =2\pi /y_0$

Answer 1

The correct option is B $\lambda =\pi y_0/2$

Wave velocity$v=\frac{coefficientoft}{coefficientofx}=\frac{2\pi f}{2\pi /\lambda }=\lambda f$
Maximum particle velocity $v_{max}=\omega A=2\pi f{y}_0$
given, $v_{max}=4v$
$\Rightarrow 2\pi f{y}_0=4f\Rightarrow \lambda =\frac{\pi y_0}{2}$