A transverse sinusoidal wave is moving along string in the positive direction of an xaxis with a speed of 80 m/s. At t = 0, the string particle atx = 0 has a transverse displacement of 4.0 cmfrom its equilibrium position and is not moving. The maximum transverse speed of the string particle at x = 0is 16 m/s. (a)What is the frequency of the wave? (b)What is the wavelength of the wave?If $\mathit{y}\left(x,t\right)\mathbf{=}{\mathit{y}}_{\mathbf{m}}\mathit{s}\mathit{i}\mathit{n}\left(kx\pm \omega t+\varphi \right)$is the form of the wave equation, (a)What is${\mathit{y}}_{\mathbf{m}}$, (b)What is k, (c)What is $\mathit{\omega }$, (d)What is, and (e)What is the correct choice of sign in front of$\mathit{\omega }$?

• The speed of the wave, v = 80 m/s .
• At x = 0 and t = 0, the transverse displacement,$y_m=4.0cmor0.04m$ .
• The maximum transverse speed of the string particle at, x = 0 , u = 16 m/s .

We use the standard equation of transverse sinusoidal wave moving in a positive direction of the x-axis and determine the required quantities.

Formula:

The angular velocity of the wave,$u=\omega y_m$ (i)

The angular frequency of the wave,$\omega =2\mathrm{\pi f}$ (ii)

The velocity of the wave,$v=\lambda f$ (iii)

The wavenumber of the wave,$k=\frac{2\mathrm{\pi }}{\lambda }$ (iv)

The expression of the wave, $y=y_m\sin \left(kx-\omega t+\varphi \right)$ (v)

It is given thatat x = 0 and t = 0 , the transverse displacement is 4.0 cm , and the particle is not moving. Thus, it indicates that this displacement is maximum displacement$\left(y_m\right)$ ( of the particle. Using the formula for transverse speed, we can determine the value ofangular speed,$\omega$ and then frequency, f .

From equation (i), the angular frequency can be given as:

$$\begin{gathered} \omega =\frac{u}{y_m} \\ =\frac{16}{0.04} \\ =400rad/s \end{gathered}$$

Now, equating equation (ii) and the given values, the frequency of the wave is given as:

$$\begin{gathered} f=\frac{400rad/s}{2\times 3.14} \\ =64Hz \end{gathered}$$

Hence, the value of the frequency is 64 Hz .

Using equation (iii) and the given values, we get the wavelength of the wave is given as:

$$\begin{gathered} \lambda =\frac{v}{f} \\ =\frac{80}{64} \\ =1.3m \end{gathered}$$

Hence, the value of the wavelength is 1.3 m .

It is given that at x = 0 and t = 0, the transverse displacement = 4.0 cm and the particle is not moving. Thus, it indicates that this displacement is the maximum displacement $\left(y_m\right)$of the particle. Hence$y_m=4.0cmor0.04m$ .

Hence, the value of maximum amplitude is 0.04 m .

Using equation (iv) and the given values, we get the wave vector of the wave is given as:

$$\begin{gathered} K=\frac{2\times 3.14}{1.3} \\ =4.8 \\ \approx 5.0m^{-1} \end{gathered}$$

Hence, the value of the wave vector is$5.0m^{-1}$ .

The angular frequency is calculated using the transverse speed of the particle. It is already calculated in part (a). Hence, using equation (i), we get the angular frequency as:

$$\begin{gathered} \omega =\frac{16}{0.04} \\ =400rad/s \end{gathered}$$

Hence, the value of wave vector is 400 rad/s .

To determine the phase angle, we use the standard equation (v) of the wave, we get the phase angle as follows:

At x =0 and t = 0,

$$\begin{gathered} y=y_m \\ =0.04m \\ 0.04=0.04\sin \left(k\left(0\right)+\omega \left(0\right)+\varphi \right) \\ 0.04=0.04\sin \left(\varphi \right) \\ \sin \varphi =1 \\ \varphi =\frac{\mathrm{\pi }}{2} \end{gathered}$$

Hence the phase angle is$\varphi =\frac{\mathrm{\pi }}{2}$ .

The sign of $\omega$should be negative as the wave is moving along the positive direction of x axis.