In a sine wave,what is the minimum distance b | vedclass

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(A) In a sine wave,the displacement of a particle is given by $y = A \sin(kx - \omega t)$.The particle velocity is $v_p = \frac{\partial y}{\partial t

Vedclass · Accepted Answer

$\frac{\lambda}{2}$

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(A) In a sine wave,the displacement of a particle is given by $y = A \sin(kx - \omega t)$.The particle velocity is $v_p = \frac{\partial y}{\partial t} = -A\omega \cos(kx - \omega t)$.The speed of the particle is $|v_p| = |A\omega \cos(kx - \omega t)|$.Two particles have the same speed if their displacement magnitudes are equal,i.e.,$|y_1| = |y_2|$.For a sine wave,points with the same speed are those that have the same magnitude of displacement from the mean position.Specifically,the points at the crest $(A)$ and the trough $(B)$ both have zero velocity (speed = $0$). The distance between a crest and the adjacent trough is $\frac{\lambda}{2}$.Alternatively,any two points symmetric about the mean position or the extremes will have the same speed. The minimum distance between two such points is $\frac{\lambda}{2}$.

In a sine wave,what is the minimum distance between two particles that always have the same speed?

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