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(A) The equation of a simple harmonic wave is given by $y = a \sin \frac{2 \pi}{\lambda} (vt - x)$.By differentiating the displacement $y$ with respec

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$\frac{\lambda}{4 \pi}$

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(A) The equation of a simple harmonic wave is given by $y = a \sin \frac{2 \pi}{\lambda} (vt - x)$.By differentiating the displacement $y$ with respect to time $t$,we obtain the particle velocity:$v_p = \frac{dy}{dt} = a \cdot \frac{2 \pi v}{\lambda} \cos \frac{2 \pi}{\lambda} (vt - x)$.The maximum particle velocity $(v_{p, \max})$ occurs when the cosine term is $1$:$v_{p, \max} = \frac{2 \pi v a}{\lambda}$.According to the problem,the maximum particle velocity is half the wave velocity $(v)$:$v_{p, \max} = \frac{v}{2}$.Equating the two expressions for $v_{p, \max}$:$\frac{v}{2} = \frac{2 \pi v a}{\lambda}$.Solving for the amplitude $a$:$a = \frac{\lambda}{4 \pi}$.

The maximum particle velocity in a wave motion is half the wave velocity. Then the amplitude of the wave is equal to

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