Question

A sinusoidal transverse wave is traveling along a string in the negative direction of an xaxis. Figure 16-25 shows a plot of the displacement as a function of position at time t=0; the scale of the y axis is set by ${\mathit{y}}_{\mathbf{s}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}$. The string tension is 3.6N, and its linear density is 25 g/m. (a) Find the amplitude, (b) Find the wavelength, (c) Find the wave speed, and (d) Find the period of the wave. (e) Find the maximum transverse speed of a particle in the string. If the wave is of the form $\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}{\mathit{y}}_{\mathbf{m}}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{k}\mathbf{x}\mathbf{\pm }\mathbf{\omega }\mathbf{t}\mathbf{+}\mathbf{f}\mathbf{)}$, (f) What is ,(g) What is $\left(\omega \right)$, (h) What is $\left(\varphi \right)$, and (i) What is the correct choice of sign in front of ?

Short answer

1. The amplitude of the wave is 5 cm or 0.05 m
2. The wavelength is 40 cm
3. The wave speed is 12 m/s
4. Period of the wave is 0.033 s
5. The maximum speed of a particle in the string is 9.4 m/s
6. Angular wave number of the wave is 16 /m
7. Angular frequency of the wave is $1.9\times {10}^2{s}^{-1}$
8. Phase difference of the wave is 0.93 rad
9. The sign in front of $\omega$is positive

Step-by-step solution

The given data

• The scale ofy axis is set at ${\mathrm{y}}_{\mathrm{s}}=4.0\mathrm{cm}\mathrm{or}0.04\mathrm{m}$
• String tension, T=3.6 N
• Linear density of the string,$\mathrm{\mu }=25\mathrm{g}/\mathrm{cm}\mathrm{or}0.025\mathrm{kg}/\mathrm{m}$

Understanding the concept of wave equation

The maximum displacement of the particle undergoing oscillation, perpendicular to the direction of motion, is known as the amplitude of wave. The distance traveled by the wave along the axis of motion in one complete revolution, is known as wavelength of wave. The time taken by the wave to complete one oscillation is called Time Period.

Formula:

The general expression of the wave, $\mathrm{y}(\mathrm{x},\mathrm{t})={\mathrm{y}}_{\mathrm{m}}\sin \left(\mathrm{kx}\pm \mathrm{\omega t}+\mathrm{\phi }\right)$ (i)

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

The wave number of a wave, $k=\frac{2\mathrm{\pi }}{\lambda }$ (iii)

The speed of the transverse wave, $v=\sqrt{\left(\frac{T}{\mu }\right)}$ (iv)

The period of a wave, $t=\frac{\lambda }{v}$ (v)

Here, T is the tension in string, $\lambda$is the wavelength and v is the speed of the wave.

a) Calculation for the amplitude of the wave

From the graph, we can observe that the maximum displacement of the particle along y axis is the amplitude of the wave. Hence, the value of amplitude is 5 cm or 0.05 m

b) Calculation of the wavelength

From the graph, by measuring the distance between two consecutive crests, we get the wavelength to be 40 cm

c) Calculation of speed of the wave

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

$$\begin{gathered} v=\sqrt{\frac{3.6N}{0.025}} \\ =\sqrt{144} \\ =12m/s \end{gathered}$$

Hence, the wave speed of the wave is 12 m/s

d) Calculation of the time period

Using equation (v) and the given values, we get the time period as:

$$\begin{gathered} t=\frac{0.4m}{12m/s} \\ =0.033s \end{gathered}$$

Hence, the time period of the wave is 0.033 s

e) Calculation of the maximum velocity

The maximum transverse speed of a particle in the string can be calculated as:

$$\begin{gathered} v_y=\frac{dy}{dt} \\ =\frac{d}{dt}\left[y_m\sin \left(kx\pm \omega t+\phi \right)\right] \\ =\pm \omega y_m\cos \left(kx\pm \omega t+\phi \right) \end{gathered}$$

At maximum velocity,$\left(kx\pm \omega t+\phi \right)=1$

$v_{y}max=\omega y_m$

Using equation (ii) in the above equation of maximum velocity, we get

$$\begin{gathered} v_{ymax}=\frac{2\mathrm{\pi }}{t}{y}_m \\ =\frac{2\times 3.14}{0.033s}\times 0.05m \\ =9.4m/s \end{gathered}$$.

Hence, the value of transverse speed of a particle is 9.4 m/s

f) Calculation of wave number

Using equation (ii) and the given values, the wave number is given as:

$\begin{array}{l}k=\frac{2\times 3.14}{0.4\mathrm{m}}\\ =15.7/m\\ \approx 16/m\end{array}$

Hence, the wave number is 16/m

g) Calculation of angular frequency

Using equation (ii) and the given values, we get the angular frequency of the wave is given as:

$\begin{array}{l}\omega =\frac{2\times 3.14}{0.033}\\ =190/s\\ =1.9\times {10}^2{s}^{-1}\end{array}$

Hence, the angular frequency is$1.9\times {10}^2{s}^{-1}$

h) Calculation of the phase difference

At t = 0, x = 0,the expression of the wave becomes:

$y=y_m\sin \left(\phi \right)$

It is mentioned that the y axis is set at$y\left(y_s\right)=0.04mandwecalculated{y}_m=0.05m$

Hence, the value of phase difference is given as:

$$\begin{gathered} \sin \phi =\frac{y_s}{y_m} \\ \sin \phi =\frac{0.04m}{0.05m} \\ \sin \phi =0.8 \\ \phi ={\sin }^{-1}\left(0.8\right) \\ =0.93rad \end{gathered}$$

Phase difference is 0.93 rad.

i) Finding the sign of the angular frequency

Sign in front of the angular frequency $\omega$ is positive.