Question

A sine wave has an amplitude A and wavelength $\lambda$. The ratio of particle velocity and the wave velocity is equal to ($2\pi A=\lambda$ )

• A. $\ge 1$
• B. $=1$
• C. $\le 1$
• D. None of these

Answer 1

The correct option is B $=1$

Amplitude $=A$

Wavelength $=\lambda$

Particle velocity $=\omega \sqrt{A^2-x^2}\Rightarrow V_{max}=\omega A$

Wave velocity $=\frac{\omega }{k}=\frac{\omega \lambda }{2\pi }$

$\therefore Ratio=\frac{V_{max}}{V_{\omega }}=\frac{\omega A}{\omega \lambda /2\pi }=\frac{2\pi A}{\lambda }$

$\therefore$ If $\frac{2\pi A}{\lambda }=1$

Then only, $2\pi A=\lambda$

Hence, option $B$ is correct.