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Q11P

Expert-verifiedFound in: Page 472

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**

a) At *t*=0, a plot of *y* vs *x* in 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 .

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

**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**

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 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.

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.

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.

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.

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

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

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

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

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

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:

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

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

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:

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

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

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