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(B) The distance between a crest and the neighbouring trough at the same instant is half the wavelength $(\lambda/2)$.Given $\lambda/2 = 4.0 \, cm$,so

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$\frac{5\pi}{2} \, cm/s$

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(B) The distance between a crest and the neighbouring trough at the same instant is half the wavelength $(\lambda/2)$.Given $\lambda/2 = 4.0 \, cm$,so $\lambda = 8.0 \, cm$.The distance between a crest and a trough at the same place is twice the amplitude $(2a)$.Given $2a = 1.0 \, cm$,so $a = 0.5 \, cm$.The time interval between two consecutive crests at the same place is the time period $(T)$.Given $T = 0.4 \, s$.The angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.4} = 5\pi \, rad/s$.The maximum speed of the vibrating particles is given by $v_{max} = a\omega$.Substituting the values: $v_{max} = 0.5 	imes 5\pi = 2.5\pi = \frac{5\pi}{2} \, cm/s$.

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