Question

A sinusoidal transverse wave of wavelength 20cmtravels along a string in the positive direction of anaxis. The displacement y of the string particle at x=0is given in Figure 16-34 as a function of time t. The scale of the vertical axis is set by${\mathit{y}}_{\mathbf{s}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{cm}$The wave equation is to be in the form$\mathbf{y}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}{\mathbf{y}}_{\mathbf{m}}\mathbf{sin}\mathbf{(}\mathbf{kx}\mathbf{\pm }\mathbf{\omega t}\mathbf{+}\mathbf{\varphi }\mathbf{)}$. (a) At t=0, is a plot of y versus x in the shape of a positive sine function or a negative sine function? (b) What is${\mathit{y}}_{\mathbf{m}}$, (c) What isk,(d) What is$\mathit{\omega }$, (e) What is$\mathit{\phi }$ (f) What is the sign in front of$\mathit{\omega }$, and (g) What is the speed of the wave? (h) What is the transverse velocity of the particle at x=0when t=5.0 s?

Short answer

a) At t=0, a plot ofy vs xin the slope of a negative sine function as:$y(x,0)=-y_{m}sin\left(kx\right)$ .

b) The amplitude $y_m$is 4.0 cm.

c) The angular wave number k is,0.31rad/cm .

d) The angular frequency $\omega$is,0.63 rad/s .

e) The phase constant$\mathrm{\varphi }$ is,$\mathrm{\pi }$ .

f) The sign in front of $\omega$is, negative.

g) The speed of the wave v is,2.0 cm/s.

h) The transverse velocity of the particle at x=0 when t=5.0 s is,-2.5 cm/s .

Step-by-step solution

The given data

• The wavelength of the wave,$\lambda =20\mathrm{cm}$.
• The scale of the vertical axis is set by$y_s=40cm$.
• The wave equation is to be in the form,$y(x,t)=y_{m}sin(kx\pm \omega t+\varphi )$ .

Understanding the concept of wave equation

By using a general expression for a sinusoidal wave traveling along the +xdirection and corresponding formulas, we can find the amplitude, angular wave numberk, angular frequency$\mathit{\omega }$, the phase constant$\mathbf{\varphi }$, the sign in front of$\mathbf{\omega }$, and the speed of the wave vand the transverse velocity of the particle at x=0when t=5.0 s.

Formula:

A general expression for a sinusoidal wave traveling along the +x direction,

$y(x,t)=y_{m}sin(kx\pm \omega t+\varphi )$ (i)

The angular wave number,$k=\frac{2\mathrm{\pi }}{\lambda }$ (ii)

The angular frequency,$\omega =\frac{2\mathrm{\pi }}{T}$ (iii)

The frequency,$f=\frac{1}{T}$ (iv)

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

The transverse velocity of the particle, $u\left(x,t\right)=\frac{\partial y}{\partial t}$ (vi)

a) Plotting y versus x graph

A general expression for a sinusoidal wave traveling along the direction using equation (vi) is given as:

$y(x,t)=y_{m}sin(kx\pm \omega t+\varphi )$

Figure 16-34 shows that at x=0

$y(0,t)=y_{m}sin(-\omega t+\varphi )\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots .\left(1\right)$

And it is a positive sine function. That is

$y(0,t)=+y_{m}sin\left(\omega t\right)$

For the sin function, we can write that

From equation (1) and (2), we can say that the phase constant must be

$\varphi =\pi$

At t = 0, we have

$y(x,0)=y_{m}sin(kx+\pi )$

Using equation (2), we get the displacement equation as:

$y(x,0)=-y_{m}sin\left(kx\right)$.

which is a negative sine function. A plot of $y\left(x,0\right)$is plotted below.

b) Calculation for amplitude

From the figure we see that the amplitude is

$y_m=4.0\mathrm{cm}$.

Hence, the value of amplitude of the function is 4.0 cm.

c) Calculation for the wavenumber

Using equation (ii) and the given value of wavelength, the angular wave number is given by:

$\begin{array}{l}k=\frac{2\mathrm{\pi }}{2\pi }\\ =0.31\mathrm{rad}/\mathrm{cm}\end{array}$

Hence, the value of wavenumber is 0.31 rad/cm.

d) Calculation for the angular frequency

Using equation (iii), the angular frequency is given by:

$$\begin{gathered} \omega =\frac{2\mathrm{\pi }}{10} \\ =0.63\mathrm{rad}/\mathrm{s} \end{gathered}$$

Hence, the value of the angular frequency is 0.63 rad/s.

e) Calculation for the phase constant

The figure shows that at x=0,

$y(0,t)=y_{m}sin(-\omega t+\varphi )$

And it is a positive sine function. That is

$\mathrm{y}(0,\mathrm{t})=+{\mathrm{y}}_{\mathrm{m}}\sin \left(\mathrm{\omega t}\right)$

Therefore, the phase constant must be$\varphi =\pi$.

Hence, the value of phase constant is $\mathrm{\pi }$.

f) Finding the sign of angular frequency

The sign is minus since the wave is traveling in the +x direction. Hence, the sign of the angular frequency is negative.

g) Calculation of the speed of the wave

Using equation (iv), the frequency of the wave is given as:

$$\begin{gathered} f=\frac{1}{10} \\ =0.10s^{-1} \end{gathered}$$

Therefore, using equation (v) and the above value f frequency, the speed of the wave is given as:

$$\begin{gathered} v=0.10\times 20 \\ =2.0\mathrm{cm} \end{gathered}$$

Hence, the value of speed of the wave is 2.0 cm/s.

h) Calculation of the transverse velocity

From the results above, the wave may be expressed as

$$\begin{gathered} y\left(x,t\right)=4.0\sin \left(\frac{\mathrm{\pi x}}{10}-\frac{\mathrm{\pi t}}{5}+\mathrm{\pi }\right) \\ =-4.0\sin \left(\frac{\mathrm{\pi x}}{10}-\frac{\mathrm{\pi t}}{5}\right) \end{gathered}$$

Using the equation (vi ) and the above wave equation, the transverse velocity is given as:

$$\begin{gathered} u\left(x,t\right)=\frac{d}{dt}\left(-4.0\sin \left(\frac{\mathrm{\pi x}}{10}-\frac{\mathrm{\pi t}}{5}\right)\right) \\ =-4.0\left(\frac{\mathrm{\pi }}{t}\right)\cos \left(\frac{\mathrm{\pi x}}{10}-\frac{\mathrm{\pi t}}{5}\right) \end{gathered}$$

Hence, at the required values, the value of transverses velocity is given by:

$$\begin{gathered} u\left(0,5,0\right)=-4.0\left(\frac{\mathrm{\pi }}{5.0}\right)\cos \left(-\frac{\mathrm{\pi }\times 5.0}{5}\right) \\ =-2.5\mathrm{cm}/\mathrm{s} \end{gathered}$$

Hence, the value of transverse velocity is 2.5 cm/s.