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Question

(A) Let the wave equation be $y = A \sin(\omega t - kx)$.The particle velocity is given by $v_p = \frac{\partial y}{\partial t} = A\omega \cos(\omega

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amplitude of the wave is less than $\frac{\lambda}{2\pi}$

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(A) Let the wave equation be $y = A \sin(\omega t - kx)$.The particle velocity is given by $v_p = \frac{\partial y}{\partial t} = A\omega \cos(\omega t - kx)$.The maximum particle speed is $v_{p, 	ext{max}} = A\omega$.The wave speed is $v = \frac{\omega}{k} = \frac{\omega}{2\pi / \lambda} = \frac{\omega \lambda}{2\pi}$.For the particle speed to be always less than the wave speed,the maximum particle speed must be less than the wave speed:$v_{p, 	ext{max}} $A\omega Dividing both sides by $\omega$,we get:$A < \frac{\lambda}{2\pi}$.

In a plane progressive harmonic wave,the particle speed is always less than the wave speed if:

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