A sinusoidal transverse wave traveling in the positive direction of an xaxis has amplitude of 2.0 cm , a wavelength of 10 cm , and a frequency of 400 Hz. If the wave equation is of the form y (x,t) $\mathbf{=}{\mathit{y}}_{\mathbf{m}}\mathit{s}\mathit{i}\mathit{n}\left(kx\pm \omega t\right)$, what are (a) role="math" localid="1660983337674" ${\mathit{y}}_{\mathbf{m}}$, (b) k , (c)$\mathit{\omega }$ , and (d) the correct choice of sign in front of $\mathit{\omega }$? What are (e) the maximum transverse speed of a point on the cord and (f) the speed of the wave?

1. The amplitude of the wave,${\mathrm{y}}_{\mathrm{m}}=2.0\mathrm{cm}=0.02\mathrm{m}$ .
2. The wavelength of the wave,$\mathrm{\lambda }=10\mathrm{cm}=0.1\mathrm{m}$.
3. Frequency of the wave,$\mathrm{f}=400\mathrm{Hz}$.
4. The wave is in the positive direction of the x-axis.
5. The form of the wave equation,$y\left(x,t\right)=y_m\sin \left(kx\pm \omega t\right)$

The amplitude of the wave is given by the maximum displacement of the wave.

The angular frequency is the number of oscillations or the angular displacement of the wave within the given period of the oscillation that is the reverse of its frequency value.

The maximum transverses speed of the wave is given by the derivative of the displacement with time.

Thus, using this concept, the maximum amplitude after calculating the derivation of the displacement equation gives us the maximum transverse speed. Using the given concept and the data, the required values are calculated.

Formulae:

The wavenumber of the wave,

$k=\frac{2\mathrm{\pi }}{\lambda }$ (i)

The angular frequency of the wave,

$\omega =2\pi f$ (ii)

The transverse speed of the wave,

$v_m=\omega y_m$ (iii)

The speed of the wave,

$v=\lambda f$ (iv)

Here f is the frequency of the wave and $y_m$is the amplitude of the wave.

The amplitude of the wave is given as:

$y_{\mathit{m}}\mathit{=}\mathit{0}\mathit{.}\mathit{002}\mathit{}\mathrm{m}$

Hence, the amplitude is0.002 m .

The wavenumber of the wave can be calculated from equation (i) as follows:

$\begin{array}{l}k=\frac{2\mathrm{\pi }}{01\mathrm{m}}\\ =62.{8\mathrm{m}}^{-1}\\ ~63{\mathrm{m}}^{-1}\end{array}$

Hence, the value of the wavenumber is$63{\mathrm{m}}^{-1}$ .

The angular frequency of the wave can be calculated from equation (ii) as follows:

$$\begin{gathered} \omega =2\pi \left(400\mathrm{Hz}\right) \\ =2514.2\mathrm{rad}/\mathrm{s} \\ =2.5\times {10}^3\mathrm{rad}/\mathrm{s} \end{gathered}$$

Hence, the value of the angular frequency is$2.5\times {10}^3\mathrm{rad}/\mathrm{s}$ .

A minus sign will be used before the$\omega t$term in the equatioon of the sine function because the wave is traveling in the positivexdirection.

The wave equation using the given values can be written as:

$y\left(x,t\right)=\left(0.02m\right)\sin \left(\left(63{\mathrm{m}}^{-1}\right)x-\left(2.5\times {10}^3{\mathrm{s}}^{-1}\right)t\right)$

Hence, the correct sign of the angular frequency is minus.

The maximum transverse speed of a point on the cord can be calculated from equation (iii) as follows:

$$\begin{gathered} V_m=(2510\mathrm{rad}/\mathrm{s})(0.02\mathrm{m}) \\ =50.2\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of the maximum speed is 50.2 m/s.

Substituting the values in equation (iv), the speed of the wave can be calculated as,

$$\begin{gathered} v=\left(0.1\mathrm{m}\right)\left(400\mathrm{Hz}\right) \\ =40\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of the speed is 40 m/s .