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

Answer: (C)

2$\pi$A

Answer 2

The given wave equations is

$y = A\sin(\omega t - kx)$

Wave velocity, $v = \frac{\omega}{k}$ …. (i)

Particle velocity,

$v_{p} = \frac{dy}{dt} = A\omega\cos(\omega t - kx)$

Maximum particle velocity,

$(v_{P})_{\max}$ …. (ii)

According to the given questions

$v = (v_{p})_{\max}$

$\frac{\omega}{k} = A\omega$ (Using (i) and (ii))

$\frac{1}{k} = Aor\frac{\lambda}{2\pi} = A$ $\left( \because k = \frac{2\pi}{\lambda} \right)$

$\lambda = 2\pi A$