• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q11P

Expert-verified Found in: Page 472 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # A sinusoidal transverse wave of wavelength 20cm travels along a string in the positive direction of an axis. The displacement y of the string particle at x=0 is given in Figure 16-34 as a function of time t. The scale of the vertical axis is set by ${{\mathbit{y}}}_{{\mathbf{s}}}{\mathbf{=}}{\mathbf{4}}{\mathbf{.}}{\mathbf{0}}{\mathbf{cm}}$ The wave equation is to be in the form ${\mathbf{y}}{\mathbf{\left(}}{\mathbf{x}}{\mathbf{,}}{\mathbf{t}}{\mathbf{\right)}}{\mathbf{=}}{{\mathbf{y}}}_{{\mathbf{m}}}{\mathbf{sin}}{\mathbf{\left(}}{\mathbf{kx}}{\mathbf{±}}{\mathbf{\omega t}}{\mathbf{+}}{\mathbf{\varphi }}{\mathbf{\right)}}$. (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 ${{\mathbit{y}}}_{{\mathbf{m}}}$, (c) What is k,(d) What is ${\mathbit{\omega }}$, (e) What is ${\mathbit{\phi }}$ (f) What is the sign in front of ${\mathbit{\omega }}$, and (g) What is the speed of the wave? (h) What is the transverse velocity of the particle at x=0 when t=5.0 s? a) At t=0, a plot of y vs x in the slope of a negative sine function as:$y\left(x,0\right)=-{y}_{m}sin\left(kx\right)$ .

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 .

See the step by step solution

## Step 1: 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,$y\left(x,t\right)={y}_{m}sin\left(kx±\omega t+\varphi \right)$ .

## Step 2: Understanding the concept of wave equation

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

Formula:

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

$y\left(x,t\right)={y}_{m}sin\left(kx±\omega t+\varphi \right)$ (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)

## Step 3: a) Plotting y versus x graph

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

$y\left(x,t\right)={y}_{m}sin\left(kx±\omega t+\varphi \right)$

Figure 16-34 shows that at x=0

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

And it is a positive sine function. That is

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

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

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

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

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

which is a negative sine function. A plot of $y\left(x,0\right)$is plotted below. ## Step 4: 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.

## Step 5: 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.

## Step 6: d) Calculation for the angular frequency

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

$\omega =\frac{2\mathrm{\pi }}{10}\phantom{\rule{0ex}{0ex}}=0.63\mathrm{rad}/\mathrm{s}$

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

## Step 7: e) Calculation for the phase constant

The figure shows that at x=0,

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

And it is a positive sine function. That is

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

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

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

## Step 8: 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.

## Step 9: g) Calculation of the speed of the wave

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

$f=\frac{1}{10}\phantom{\rule{0ex}{0ex}}=0.10{s}^{-1}$

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

$v=0.10×20\phantom{\rule{0ex}{0ex}}=2.0\mathrm{cm}$

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

## Step 10: h) Calculation of the transverse velocity

From the results above, the wave may be expressed as

$y\left(x,t\right)=4.0\mathrm{sin}\left(\frac{\mathrm{\pi x}}{10}-\frac{\mathrm{\pi t}}{5}+\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}=-4.0\mathrm{sin}\left(\frac{\mathrm{\pi x}}{10}-\frac{\mathrm{\pi t}}{5}\right)$

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

$u\left(x,t\right)=\frac{d}{dt}\left(-4.0\mathrm{sin}\left(\frac{\mathrm{\pi x}}{10}-\frac{\mathrm{\pi t}}{5}\right)\right)\phantom{\rule{0ex}{0ex}}=-4.0\left(\frac{\mathrm{\pi }}{t}\right)\mathrm{cos}\left(\frac{\mathrm{\pi x}}{10}-\frac{\mathrm{\pi t}}{5}\right)$

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

$u\left(0,5,0\right)=-4.0\left(\frac{\mathrm{\pi }}{5.0}\right)\mathrm{cos}\left(-\frac{\mathrm{\pi }×5.0}{5}\right)\phantom{\rule{0ex}{0ex}}=-2.5\mathrm{cm}/\mathrm{s}$

Hence, the value of transverse velocity is 2.5 cm/s. ### Want to see more solutions like these? 