Question

Show that the particle speed can never be equal to the wave speed in a sine wave if the amplitude is less than wavelength divided by 2π.

Answer 1

Equation of the wave is given by 
\[y = A\sin\left( \omega t - kx \right)\] 
where
          A is the amplitude
           ω is the angular frequency
           k is the wave number
Velocity of wave, \[y = A\sin\left( \omega t - kx \right)\]
Velocity of particle, \[v_p = \frac{dy}{dt} = A\omega \cos\left( \omega t - kx \right)\]
Max velocity of particle, 
\[v_{p_\max}  = A\omega\]
As given
\[A < \frac{\lambda}{2\pi}\]

\[v_{p_\max}  = \frac{\lambda\omega}{2\pi}\] 

\[ v_{p_\max}  < \frac{\omega}{k}                          \left[ \because \frac{2\pi}{\lambda} = k \right]\]