Question

A transverse wave is expressed as, $y=y_0\sin 2\pi ft$ . For what value of $\lambda .$ maximum particle velocity equals to 4 times the wave velocity?

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

Answer 1

Answer: (A)

Particle velocity is the velocity $v$ of a particle in a medium as it transmits a wave.
When $A$ is amplitude and co is angular speed, then
$v=A\omega$... (i)
Phase velocity of a wave is the rate at which the phase of the wave propagates in space.
This is the velocity at which the phase of any one frequency component of the wave will propagate.
It is given by $v_p=\frac{\omega }{k}$ ... (ii)
where $k$ is a wave vector Given equation is .
$y=y_0\sin 2\pi ft$ where $A=y_0,k=\frac{2\pi }{\lambda }$
According to question, $v=4v_p$
$\therefore A\omega =4\frac{\omega }{k}$
$\therefore y_0\omega =\frac{4\omega }{k}$
$\Rightarrow y_0=\frac{4}{k}=\frac{4}{2\pi }\times \lambda$
Therefore, wavelength, $\lambda =\frac{\pi y_0}{2}$