91Ó°ÊÓ

The wave equation is,$y=(2.0\mathrm{mm})\sin \left((20{\mathrm{m}}^{-1})\mathrm{x}-\left(600{\mathrm{s}}^{-1}\right)\mathrm{t}\right)$

We can find the values of amplitude, angular frequency, and wave constant by comparing the given equation with the general equation of the wave. Using them, we can find the frequency, wavelength, and velocity of the wave and maximum speed of the particle on the string using corresponding relations.

Formulae:

The wavenumber of the wave,

$k=\frac{2\mathrm{π}}{\mathrm{λ}}$ (i)

The angular frequency of the wave,

$\mathrm{Ó\neg }=2\mathrm{Ï€´Ú}$ (ii)

The transverse speed of the wave,

$v_m=\mathrm{Ó\neg }{y}_m$ (iii)

The speed of the wave,

$v=\mathrm{λ}f$ (iv)

Here Here f is the frequency of the wave $\mathrm{λ}$is the wavelength of the wave and $y_m$is the amplitude of the wave.

The general equation of a translational wave on a string is,

$y=y_m\sin \left(kx-\mathrm{Ó\neg }t\right)$ (1)

The given equation of the wave is,

$y=(2.0\mathrm{mm})\sin \left((20{\mathrm{m}}^{-1})\mathrm{x}-\left(600{\mathrm{s}}^{-1}\right)\mathrm{t}\right)$ (2)

Comparing equations (1) and (2), we can find the value of amplitude as:

$y_m=2.0\mathrm{mm}$

Therefore, the amplitude of a wave traveling along a string is 2.0 mm.

Comparing equations (1) and (2), we can find the angular frequency as:

$\mathrm{Ó\neg }=600{\mathrm{s}}^{-1}$

From equation (i) and the given values, the frequency of the wave can be calculated as,

$$\begin{gathered} \mathrm{f}=\frac{600}{2\mathrm{Ï€}} \\ =95.45 \\ â‰\dots 95\mathrm{Hz} \end{gathered}$$

Therefore, the frequency of a wave traveling along a string is 95 Hz.

Comparing equations (1) and (2), we can find the wavenumber as:

$k=20{\mathrm{m}}^{-1}$

Substitute the values in equation (ii), and the wavelength can be calculated as:

$$\begin{gathered} \mathrm{λ}=\frac{2\mathrm{π}}{20} \\ =0.314m \end{gathered}$$

Substitute the values in equation (iii), and we can calculate the velocity of the wave as:

$$\begin{gathered} \mathrm{v}=95.54\left(0.314\right) \\ =30\mathrm{m}/\mathrm{s} \end{gathered}$$

Since the wave is traveling in the +x direction, velocity is in the +x direction.

Therefore, the velocity (including sign) of a wave traveling along a string is +30 m/s.

From part (c), the wavelength we calculated is:

$$\begin{gathered} \mathrm{λ}=0.314 \\ â‰\dots 0.31\mathrm{m} \\ =31\mathrm{cm} \end{gathered}$$

Therefore, the wavelength of a wave traveling along a string is 31 cm.

Substitute the values in equation (iv), and the maximum speed of a wave can be calculated as:

$$\begin{gathered} {\mathrm{v}}_{\mathrm{m}}=0.002\left(600\right) \\ =1.2\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the maximum transverse speed of a particle in the string is 1.2 m/s.