Answer

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**Hint:**We need to compare each of the wave equations given in the question to the standard wave equation. Then the wave speed will be given by the ratio of $\omega $ and $k$, and the maximum transverse string speed will be the product of the amplitude and $\omega $. Comparing the values for each wave we can find which option is correct.

**Formula Used:**In this solution we will be using the following formula,

$\Rightarrow v = \dfrac{\omega }{k}$

where $v$ is the wave speed, $\omega $ is the angular frequency and $k$ is the wave number.

$\Rightarrow {u_{\max }} = A\omega $

where ${u_{\max }}$ is the maximum transverse string speed, $A$ is the amplitude.

**Complete step by step answer**

To solve this problem, we need to first compare all the wave equations to the standard wave equation given as

$\Rightarrow y = A\sin \left( {kx - \omega t} \right)$

Now for a wave, the wave speed is given as,

$\Rightarrow v = \dfrac{\omega }{k}$

and the maximum transverse speed is calculated by taking the derivative of the equation with respect to time,

$\Rightarrow {u_{\max }} = {\left. {\dfrac{{dy}}{{dt}}} \right|_{\max }} = {\left. {\dfrac{d}{{dt}}\left[ {A\sin \left( {kx - \omega t} \right)} \right]} \right|_{\max }}$

So we get the value as ${u_{\max }} = A\omega $

Now for the first wave we have $y\left( {x,t} \right) = \left( {2cm} \right)\sin \left( {3x - 6t} \right)$

On comparing with $y = A\sin \left( {kx - \omega t} \right)$ we have,

$\Rightarrow A = 2cm$, $k = 3$ and $\omega = 6$

So we get for the first wave,

$\Rightarrow {v_1} = \dfrac{\omega }{k} = \dfrac{6}{3} = 2cm/s$

and ${u_{\max 1}} = A\omega = 2 \times 6 = 12cm/s$

For the second wave we have $y\left( {x,t} \right) = \left( {3cm} \right)\sin \left( {4x - 12t} \right)$

On comparing with $y = A\sin \left( {kx - \omega t} \right)$ we have,

$\Rightarrow A = 3cm$, $k = 4$ and $\omega = 12$

So we get for the second wave,

$\Rightarrow {v_2} = \dfrac{\omega }{k} = \dfrac{{12}}{4} = 3cm/s$

and ${u_{\max 2}} = A\omega = 3 \times 12 = 36cm/s$

For the third wave we have $y\left( {x,t} \right) = \left( {4cm} \right)\sin \left( {5x - 11t} \right)$

On comparing with $y = A\sin \left( {kx - \omega t} \right)$ we have,

$\Rightarrow A = 4cm$, $k = 5$ and $\omega = 11$

So we get for the third wave,

$\Rightarrow {v_3} = \dfrac{\omega }{k} = \dfrac{{11}}{5} = 2.2cm/s$

and ${u_{\max 3}} = A\omega = 4 \times 11 = 44cm/s$

Hence we can say that wave 2 has the greatest wave speed and wave 3 has the greatest maximum transverse string speed.

**Note**

The wave speed is the phase velocity of a wave. It is the velocity with which any one frequency component of the wave travels. The phase velocity of the wave can also be written in the terms of the wavelength and time period as $v = \dfrac{\lambda }{T}$.

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