The function$\mathbf{y}\left(x,t\right)\mathbf{=}\left(15.0\mathrm{cm}\right)\mathbf{}\mathbf{cos}\mathbf{}\left(\mathrm{\tau \tau x}-15\mathrm{\tau \tau t}\right)$, with x in meters and t in seconds, describes a wave on a taut string. What is the transverse speed for a point on the string at an instant when that point has the displacement $\mathit{y}\mathbf{=}\mathbf{+}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}$?

The speed of the oscillating particle perpendicular to the direction of motion of the wave is known as the transverse velocity of that wave.

Formula:

The transverse speed of the wave, $u=\frac{dy}{dt}$$u=\frac{dy}{dt}$ (i)

Using equation (i), the transverse speed of the wave is given as:

$$\begin{gathered} u=\frac{d}{dt}\left(\left(15.0cm\right)\cos \left(\mathrm{\pi x}-15\mathrm{\pi t}\right)\right) \\ =\left(225\mathrm{\pi }\mathrm{cm}\right)\sin \left(\mathrm{\pi x}-15\mathrm{\pi t}\right).......................\left(a\right) \end{gathered}$$

Squaring equation (a) and adding it to the square of, we get

role="math" localid="1660978815151" $$\begin{gathered} u^2+{\left(15\mathrm{\pi y}\right)}^2=\left[\left(225\mathrm{\pi }\mathrm{cm}\right)\sin {\left(\mathrm{\pi x}-15\mathrm{\pi t}\right)}^2\right]+{\left[15\mathrm{\pi }\times \left(15.0\mathrm{cm}\right)\cos \left(\mathrm{\pi x}-15\mathrm{\pi t}\right)\right]}^2 \\ {u}^2+{\left(15\mathrm{\pi y}\right)}^2={\left(225\mathrm{\pi }\mathrm{cm}\right)}^2\left[{\sin }^2\left(\mathrm{\pi x}-15\mathrm{\pi t}\right)+{\cos }^2\left(\mathrm{\pi x}-15\mathrm{\pi t}\right)\right] \\ {u}^2+{\left(15\mathrm{\pi y}\right)}^2={\left(225\mathrm{\pi }\mathrm{cm}\right)}^2 \end{gathered}$$

So that, the equation of transverse velocity is given as:

$$\begin{gathered} u=\sqrt{{\left(225\mathrm{\pi }\mathrm{cm}\right)}^2-{\left(15\mathrm{\pi y}\right)}^2} \\ =15\mathrm{\pi }\sqrt{\left(15{\mathrm{cm}}^2\right)-{\mathrm{y}}^2}....................\left(\mathrm{b}\right) \end{gathered}$$

Therefore, using in equation (b), we get,

$$\begin{gathered} u=15\mathrm{\pi }\sqrt{\left(15{\mathrm{cm}}^2\right)-\left(12{\mathrm{cm}}^2\right)} \\ =\pm 135\mathrm{\pi }\mathrm{cm}/\mathrm{s} \end{gathered}$$

As speed cannot be negative-

$$\begin{gathered} u=424cm/s \\ =4.24m/s \end{gathered}$$

Hence, the value of transverse speed is $4.24m/s$.