A sinusoidal wave is traveling on a string with speed 40 cm/s. The displacement of the particles of the string at x = 10 cmvaries with time according to y = (5.0 cm) $\mathbf{sin}\left[1.0-\left(4.0s^{-1}\right)t\right]$.The linear density of the string is 4.0 g/cm. (a)What is the frequency and (b) what is the wavelength of the wave? If the wave equation is of the form,$\mathbf{y}\left(x,t\right)\mathbf{=}{\mathbf{y}}_{\mathbf{m}}\mathbf{sin}\left(\mathrm{kx}\pm \mathrm{\omega t}\right)$ (c) What is${\mathbf{y}}_{\mathbf{m}}$, (d) What is k, (e) What is$\mathbf{\omega }$, and (f) What is the correct choice of sign in front of$\mathbf{\omega }$? (g)What is the tension in the string?

• Speed of the wave, v = 40 cm/s or 0.4 m/s
• The displacement of the particle x = 10 cm or 0.1 m is given by$\mathrm{y}=\left(5.0\mathrm{cm}\right)\sin \left(1.0-\left(4.0{\mathrm{s}}^{-1}\right)\mathrm{t}\right)$
• Linear density of the string,$\mu =4.0\mathrm{g}/\mathrm{cm}\mathrm{or}0.4\mathrm{kg}/\mathrm{m}$

By comparing the given equation with a general equation of a sinusoidal wave, we will calculate the required values.

Formula:

The general expression of the wave, $y(x,t)=y_m\sin (kx-wt)$ (i)

The frequency of a wave, $f=\frac{\omega }{2\pi }$ (ii)

The wavelength of a wave,$\lambda =\frac{v}{f}$ (iii)

The speed of a wave, $v=\sqrt{\frac{T}{\mu }}$ (iv)

For the wave travelling in positive x direction, at time t, displacement y for the particle located at x is given by the equation (i)

The wave equation of the given data is given by:

$$\begin{gathered} y=\left(5.0cm\right)\sin \left(1.0-\left(4.0s^{-1}\right)t\right) \\ =\left(0.5m\right)\sin \left(1.0-\left(4.0s^{-1}\right)t\right).................\left(a\right) \end{gathered}$$

Comparing it with equation (a), we will know the angular frequency is, $\omega =4rad/s$hence, the frequency of the wave is given as:

$$\begin{gathered} f=\frac{4rad/s}{2\times 3.14} \\ =0.637\mathrm{Hz} \\ \approx 0.64\mathrm{Hz} \end{gathered}$$

Hence, the value of frequency is 0.64 Hz

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

$$\begin{gathered} \mathrm{\lambda }=\frac{0.4\mathrm{m}/\mathrm{s}}{0.637{\mathrm{s}}^{-1}} \\ =0.63\mathrm{m} \end{gathered}$$

Hence, the value of the wavelength is 0.63 m

Comparing equation (a) with the given equation (i) we get amplitude,

${\mathrm{y}}_{\mathrm{m}}=0.05\mathrm{m}$

Hence, the value of the amplitude is 0.05 m

Comparing equation (a) with equation (i) we get the wave number as:

$$\begin{gathered} kx=1 \\ k=\frac{1}{x} \\ =\frac{1}{0.1} \\ =10/m \end{gathered}$$

Hence, the value of wavenumber is 10/ m

Comparing equation (a) with the equation (i), we get the angular frequency as:

$\mathrm{\omega }=4\mathrm{rad}/\mathrm{s}$

Hence, the value of angular frequency is 4 rad/s

The correct choice of the sign in front of$\omega$ will be positive x direction.

Squaring both sides of equation (iv), we get the tension formula as:

$$\begin{gathered} v^2=\frac{T}{\mu } \\ T=v^2\mu \\ ={\left(0.4\mathrm{m}/\mathrm{s}\right)}^2\times 0.4\mathrm{kg}/\mathrm{m} \\ =0.064\mathrm{N} \end{gathered}$$

Hence, the tension in the string is 0.064 N