Question

A stretched string has a mass per unit length of5.00 g/cmand a tension of 10.0N. A sinusoidal wave on this string has amplitude of 0.12mmand a frequency of 100 Hzand is traveling in the negative direction of an xaxis. If the wave equation is of the form $\mathbf{y}\left(x,t\right)\mathbf{=}{\mathbf{y}}_{\mathbf{m}}\mathbf{}\mathbf{sin}\left(\mathrm{kx}\pm \mathrm{\omega t}\right)$, (a)What is ${\mathbf{y}}_{\mathbf{m}}$, (b)What is k , (c)What is $\mathit{\omega }$, and (d)What is the correct choice of sign in front of $\mathit{\omega }$?

Short answer

a) Wave amplitude $\left(y_m\right)$is 0.120 mm

b) Wave vector (k) is $141{\mathrm{m}}^{-1}$

c) Angular frequency $\left(\omega \right)$of the wave is

d) The correct sign in front of the$\omega$ is positive.

Step-by-step solution

The given data

• Mass per unit length, $\mathrm{\mu }=5\mathrm{gm}/\mathrm{cm}$
• Tension in the string, T =10N
• Amplitude of the wave,$y_m=0.12\mathrm{mm}$
• Thewave is travelling in the negative direction of x-axis.
• Frequency of the wave, f =100 Hz

Understanding the concept of wave equation

The wave equation is of the form,

$\mathbf{y}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}{\mathbf{y}}_{\mathbf{m}}\mathbf{sin}\mathbf{(}\mathbf{kx}\mathbf{\pm }\mathbf{\omega t}\mathbf{)}\mathbf{\hspace{0.33em}}\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

where ${\mathbf{y}}_{\mathbf{m}}$ is wave amplitude, k is wave vector,$\mathit{\omega }$angular frequency We can compare the given information with the given equation in the problem to find the required quantities.

Formula:

The wavenumber of the given function,$\mathbf{k}\mathbf{=}\frac{\mathbf{2}\mathbf{\tau \tau }}{\mathbf{\lambda }}\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

The angular frequency of the function,$\mathbf{}\mathbf{\omega }\mathbf{=}\mathbf{2}\mathbf{\pi f}\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\cdot }\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

The wavelength of the wave,

$$\begin{gathered} \mathbf{\lambda }\mathbf{=}\frac{\mathbf{v}}{\mathbf{f}} \\ \mathbf{}\mathbf{}\mathbf{=}\sqrt{\frac{\mathbf{T}}{\mathbf{\mu }}}\mathbf{\times }\frac{\mathbf{1}}{\mathbf{f}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(4\right) \end{gathered}$$

a) Calculation for the amplitude of the wave

In equation $\left(1\right),y_m$represents the amplitude of the wave. It is described as the maximum displacement of the particle from its mean position, perpendicular to the direction of propagation of wave. According to the given information, the amplitude of the given wave is-

$y_m=0.12\mathrm{mm}$

b) Calculation value of wave vector

Comparing equations (2) and (4), the wave vector of a wave is given as:

$$\begin{gathered} k=2\mathrm{\pi f}\sqrt{\frac{\mathrm{\mu }}{\mathrm{T}}} \\ =2\mathrm{\pi }\left(100\mathrm{Hz}\right)\sqrt{\frac{0.5{\mathrm{kgm}}^{-1}}{10\mathrm{N}}} \\ =141{\mathrm{m}}^{-1} \end{gathered}$$

Hence, the value of wave vector is$141{\mathrm{m}}^{-1}$

c) Calculation of the angular frequency

Using equation (3), the angular frequency is given as:

$$\begin{gathered} \mathrm{\omega }=2\times 3.14\times 100\mathrm{Hz} \\ =628\mathrm{rad}/\sec \end{gathered}$$

Hence, the value of the angular frequency is 628 rad/sec

d) Finding the sign of angular frequency

Since the wave is traveling in the negative x direction, the correct sign in front of the $\omega$ is positive.