Answer

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**Hint:**Here, the equation of the transverse wave on a string is given. Here, we will first compare the given equation with the general equation of the wave. Then, we will calculate the value of the terms in each step by using the following formula.

**Formula used:**

The formula of transverse string is given by

${v_{\max }} = A\omega $

Here, ${v_{\max }}$ is the maximum speed of the wave, $A$ is the amplitude of the wave and $\omega $ is the angular frequency.

The formula of maximum transverse speed of a wave is given by

$V = \dfrac{\omega }{k}$

$V$ is the speed of the wave, $\omega $ is the angular frequency and $k$ is the wave number.

**Complete step by step answer:**

As given in the question, the equation of transverse wave is given by

$y(x,t) = \left( {2.20cm} \right)\sin \left[ {\left( {130rad\,{s^{ - 1}}} \right)t + \left( {15rad\,{m^{ - 1}}} \right)x} \right]$

Comparing the above equation with $y = A\sin \left( {\omega t - kx} \right)$ , we get

Amplitude of the wave,$A = 2.20cm = 0.022\,m$

Angular frequency of the wave, $\omega = 130rad\,{s^{ - 1}}$

Wave number, $k = 15rad\,{m^{ - 1}}$

The formula for calculating maximum transverse speed of the string is given by

${v_{\max }} = A\omega $

Putting the values, we get

${v_{\max }} = 0.022m \times 130rad\,{s^{ - 1}}$

$ \Rightarrow \,{v_{\max }} = 2.86\,m{s^{ - 1}}$

$\therefore \,{v_{\max }} \simeq 2.9\,m{s^{ - 1}}$

Therefore, the maximum transverse speed of a point in the string is $2.9\,m{s^{ - 1}}$.

**Hence, option (C) is the correct option.**

II. Now, the approximate maximum transverse acceleration can be calculated by double differentiating the equation given in the question with respect to $t$ .

Firstly, differentiating the equation, we get

${\left( {\dfrac{{\partial y}}{{\partial t}}} \right)_x} = \dfrac{\partial }{{\partial t}}\left[ {\left( {2.20\,cm} \right)\sin \left\{ {\left( {130rad\,{s^{ - 1}}} \right)t + \left( {15rad\,{s^{ - 1}}} \right)x} \right\}} \right]$

$ \Rightarrow \,{\left( {\dfrac{{\partial y}}{{\partial t}}} \right)_x} = \left( {2.20\,cm} \right)\cos \left[ {\left( {130\,rad\,{s^{ - 1}}} \right)t + \left( {15\,rad\,{m^{ - 1}}} \right)x} \right]\left( {130rad\,{s^{ - 1}}} \right)$

Again differentiating, we get

$ \Rightarrow \,{\left( {\dfrac{{{\partial ^2}y}}{{\partial {t^2}}}} \right)_x} = - \left( {2.20\,cm} \right)\sin \left[ {\left( {130\,rad\,{s^{ - 1}}} \right)t + \left( {15\,rad\,{m^{ - 1}}} \right)x} \right].{\left( {130rad\,{s^{ - 1}}} \right)^2}$

Now, we can calculate the maximum transverse acceleration by maximizing $\,{\left( {\dfrac{{{\partial ^2}y}}{{\partial {t^2}}}} \right)_x}$

For that we will take $\sin \left[ {\left( {130rad\,{s^{ - 1}}} \right)t + \left( {15rad\,{m^{ - 1}}} \right)x} \right] = 1$

$ \Rightarrow \,\left( {\dfrac{{{\partial ^2}y}}{{\partial {t^2}}}} \right) = \left( {2.20\,cm} \right) \times \left( {16900\,ra{d^2}\,{s^{ - 2}}} \right)$

$ \Rightarrow \,{\left( {\dfrac{{{\partial ^2}y}}{{\partial {t^2}}}} \right)_x} = 37180\,cm\,{s^{ - 2}}$

$ \therefore\,{\left( {\dfrac{{{\partial ^2}y}}{{\partial {t^2}}}} \right)_x} = 371.80\,m\,{s^{ - 2}}$

Therefore, approximate maximum transverse acceleration is $372\,m\,{s^{ - 2}}$

**Hence, option (B) is the correct option.**

III. Now, the approximate speed of a wave can be calculated by the following formula

Speed of wave, $V = \dfrac{\omega }{k}$

$ \Rightarrow \,V = \dfrac{{130}}{{15}}$

$ \Rightarrow \,V = 8.67\,m{s^{ - 1}}$

$ \therefore \,V \simeq \,8.7\,m{s^{ - 1}}$

Therefore, the approximate speed of the wave moving along the string is $8.7\,m{s^{ - 1}}$ .

**Hence, option (D) is the correct option.**

**Note:**Here, in the second option, you must calculate the derivative very carefully because the equation is tough to derivative. In this option, we are differentiating the equation by keeping $x$ constant. Also, remember here to change the units of amplitude into meters.

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