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**Hint:**The maximum speed of the vibrating particles can be calculated with the help of the general equation of a wave. If we differentiate the general equation of a wave, we can get the general equation of velocity.

**Complete step by step solution:**

The distance between the crest and trough is $1.0{\rm{ cm}}$ which means that the amplitude is half of $1.0{\rm{ cm}}$. This can evaluate the value of amplitude as given below,

$A$ = $\dfrac{{1{\rm{ cm}}}}{2}\\$

$A$ = $0.5\;{\rm{cm}}$

The interval at which the next crest appears is nothing but the time period, therefore, the time period of the given wave is $0.4\;{\rm{s}}$.

We can calculate the value of angular frequency with the help of time period.

$\omega = \dfrac{{2\pi }}{T}$

We will now substitute the known values in the above equation of angular frequency.

$\omega = \dfrac{{2\pi }}{{0.4\;{\rm{s}}}}\\$

$\Rightarrow$ $5\pi \;{\rm{rad/s}}$

Here, the time interval is $T$.

We know that the general equation of a wave is given as $y = A\sin( \omega t + kx)$.

So, for maximum velocity, we will differentiate the above equation with respect to time.

$v$ = $\dfrac{{dy}}{{dt}}\\$

$\Rightarrow$ $\dfrac{{d\left( {A\sin \omega t + kx} \right)}}{{dt}}\\$

$\Rightarrow$ $A\omega \left( {\cos \omega t + kx} \right)$

The maximum value of the equation is \[{v_{\max }} = A\omega \].

The equation of maximum velocity is evaluated and now we can substitute the values to get maximum velocity.

${v_{\max }} = 0.5 \times 5\pi \\$

$\Rightarrow$ $\dfrac{{5\pi }}{2}\;{\rm{cm/s}}$

Thus, the maximum speed of the vibrating particles in the given medium is calculated to be $\dfrac{{5\pi }}{2}\;{\rm{cm/s}}$.

**Thus, From the given options, only option B is correct.**

**Note:**The step in which the equation for maximum velocity comes to be \[A\omega \] is a tricky method. We should remember that the maximum value of any sine or cosine function is 1, so in order to get maximum value of the equation \[A\omega \left( {\cos \omega t + kx} \right)\], the cosine function is taken as 1.

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