A transverse wave is described by the equation $y={{y}_{o}}\sin 2\pi \left( ft-\dfrac{x}{a} \right)$. The maximum particle velocity is equal to four times the wave velocity if a is equal to:
A. $\dfrac{\pi {{y}_{o}}}{4}$
B. $\dfrac{\pi {{y}_{o}}}{2}$
C. $\pi {{y}_{o}}$
D. $2\pi {{y}_{o}}$

Hint: Wave velocity is the product of the frequency of the wave and the wavelength of the wave. The particle velocity will be given by the derivative of the wave equation. When we get the expression for both we will use the relation between them as given in the question and then find the correct answer.
Formula used:
$v=\nu \times \lambda $
${{v}_{p}}=\dfrac{dy}{dt}$

Complete answer:
In the given wave equation, $y={{y}_{o}}\sin 2\pi \left( ft-\dfrac{x}{a} \right)$, f will be the frequency of the wave and a will be equal to the wavelength of the wave. We can see that when we compare it to the standard wave equation which is $y=A\sin \left( kx+\omega t \right)$. Here k is equal to $\dfrac{2\pi }{\lambda }$and $\omega $ is equal to $2\pi \nu $. So, we get the velocity of the wave as the product of f and a. Now, to find the velocity of the particles.

We will get the velocity of the particles by taking the time derivative of the given equation.
\[{{v}_{p}}=\dfrac{dy}{dt}=\dfrac{d}{dt}\left( {{y}_{o}}\sin 2\pi \left( ft-\dfrac{x}{a} \right) \right)=2\pi f{{y}_{o}}\cos 2\pi \left( ft-\dfrac{x}{a} \right)\]. Its maximum value will be when the cos term is equal to 1. So, the maximum velocity of the particles will be \[2\pi f{{y}_{o}}\]. Now we are given that ${{v}_{p}}=4v$. When we input the values for both, we get \[2\pi f{{y}_{o}}=4fa\]. From this we will get the value of a as \[2\pi f{{y}_{o}}=4fa\Rightarrow a=\dfrac{\pi {{y}_{o}}}{2}\].

So, the correct answer is “Option B”.

Note:
When we compare the general wave equation and this wave equation, we see that there is a negative sign on the x term. This denotes the direction in which the wave is moving and as we only need the value of the wavelength, it does not affect our answer.