Question

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 $\mathit{y}\mathbf{=}\mathbf{+}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}$?

Short answer

The transverse speed for a point on the string at an instant when that point has the displacement $y=+12.0cm$ is$4.24m/s$ .

Step-by-step solution

Understanding the concept of the wave equation

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)

Calculation for the transverse speed

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 $4.24m/s$.