91Ó°ÊÓ

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:

$$\begin{gathered} u=\frac{d}{dt}\left(\left(15.0cm\right)\cos \left(\mathrm{Ï€³\ae }-15\mathrm{Ï€³Ù}\right)\right) \\ =\left(225\mathrm{Ï€}\mathrm{cm}\right)\sin \left(\mathrm{Ï€³\ae }-15\mathrm{Ï€³Ù}\right).......................\left(a\right) \end{gathered}$$

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

role="math" localid="1660978815151" $$\begin{gathered} u^2+{\left(15\mathrm{Ï€²â}\right)}^2=\left[\left(225\mathrm{Ï€}\mathrm{cm}\right)\sin {\left(\mathrm{Ï€³\ae }-15\mathrm{Ï€³Ù}\right)}^2\right]+{\left[15\mathrm{Ï€}×\left(15.0\mathrm{cm}\right)\cos \left(\mathrm{Ï€³\ae }-15\mathrm{Ï€³Ù}\right)\right]}^2 \\ {u}^2+{\left(15\mathrm{Ï€²â}\right)}^2={\left(225\mathrm{Ï€}\mathrm{cm}\right)}^2\left[{\sin }^2\left(\mathrm{Ï€³\ae }-15\mathrm{Ï€³Ù}\right)+{\cos }^2\left(\mathrm{Ï€³\ae }-15\mathrm{Ï€³Ù}\right)\right] \\ {u}^2+{\left(15\mathrm{Ï€²â}\right)}^2={\left(225\mathrm{Ï€}\mathrm{cm}\right)}^2 \end{gathered}$$

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

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

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

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

As speed cannot be negative-

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

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