Chapter 16, Problem 16.13P

(a)

To determine

The frequency of the wave.

Answer to Problem 16.13P

The frequency of the wave is $0.500\text{Hz}$.

Explanation of Solution

Given info: The wavelength of wave is $2.00\text{m}$, the amplitude is $0.100\text{m}$. The speed of wave in string is $1.00\text{m}/\text{s}$. The left end of wave is at origin at $t=0$.

The formula to calculate frequency of wave is,

$f=\frac{v}{\lambda }$

Here,

$f$ is frequency of wave.

$v$ is speed of wave.

$\lambda$ is wavelength of wave.

Substitute $1.00\text{m}/\text{s}$ for $v$ and $2.00\text{m}$ for $\lambda$ in the above expression.

$\begin{array}{c}f=\frac{1.00\text{m}/\text{s}}{2.00\text{m}}\\ =0.500\text{Hz}\end{array}$

Conclusion:

Therefore, the frequency of the wave is $0.500\text{Hz}$.

(b)

To determine

The angular frequency of the wave.

Answer to Problem 16.13P

The angular frequency of the wave is $3.14\text{rad}/\text{s}$.

Explanation of Solution

Given info: The wavelength of wave is $2.00\text{m}$, the amplitude is $0.100\text{m}$. The speed of wave in string is $1.00\text{m}/\text{s}$. The left end of wave is at origin at $t=0$.

The formula to calculate angular frequency of the wave is,

$\omega =2\pi f$

Here,

$\omega$ is angular frequency of the wave.

Substitute $0.5\text{Hz}$ for $f$ in the above expression.

$\begin{array}{c}\omega =2\pi \left(0.5\text{Hz}\right)\\ =\pi \text{rad}/\text{s}\\ =3.14\text{rad}/\text{s}\end{array}$ (1)

Conclusion:

Therefore, the angular frequency of the wave is $3.14\text{rad}/\text{s}$.

(c)

To determine

The angular wave number of the wave.

Answer to Problem 16.13P

The angular wave number of the wave is $3.14\text{rad}/\text{m}$.

Explanation of Solution

Given info: The wavelength of wave is $2.00\text{m}$, the amplitude is $0.100\text{m}$. The speed of wave in string is $1.00\text{m}/\text{s}$. The left end of wave is at origin at $t=0$.

The formula to calculate angular wave number of the wave is,

$k=\frac{2\pi }{\lambda }$

Here,

$k$ is angular wave number.

Substitute $2.00\text{m}$ for $\lambda$ in the above expression.

$\begin{array}{c}k=\frac{2\pi }{2.00\text{m}}\\ =\pi \text{rad}/\text{m}\\ =3.14\text{rad}/\text{m}\end{array}$ (2)

Conclusion:

Therefore, the angular wave number of the wave is $3.14\text{rad}/\text{m}$.

(d)

To determine

The wave function of the wave.

Answer to Problem 16.13P

The wave function of the wave is $0.100\sin \left(\pi x-\pi t\right)$.

Explanation of Solution

Given info: The wavelength of wave is $2.00\text{m}$, the amplitude is $0.100\text{m}$. The speed of wave in string is $1.00\text{m}/\text{s}$. The left end of wave is at origin at $t=0$.

The formula of standard wave equation is,

$y=A\sin \left(kx-\omega t\right)$

Here,

$A$ is amplitude of the wave.

$k$ is angular wave number of the wave.

$t$ is time period of wave.

Substitute $0.100\text{m}$ for $A$, $\pi \text{rad}/\text{m}$ for $k$ and $\pi \text{rad}/\text{s}$ for $\omega$ in the above equation.

$y=\left(0.100\text{m}\right)\sin \left(\pi x-\pi t\right)$ (3)

Conclusion:

Therefore, the function of the wave is $0.100\sin \left(\pi x-\pi t\right)$.

(e)

To determine

The equation of motion for the left end of string.

Answer to Problem 16.13P

The equation of motion for the left end of string is $0.100\sin \left(-\pi t\right)$.

Explanation of Solution

Given info: The wavelength of wave is $2.00\text{m}$, the amplitude is $0.100\text{m}$. The speed of wave in string is $1.00\text{m}/\text{s}$. The left end of wave is at origin at $t=0$.

From equation (3),

$y=\left(0.100\text{m}\right)\sin \left(\pi x-\pi t\right)$

For the left end of string the position coordinate $x$ of the wave is $0$.

Substitute $0$ for $x$ in the above equation.

$\begin{array}{c}y=\left(0.100\text{m}\right)\sin \left(\pi \left(0\right)-\pi t\right)\\ =\left(0.100\text{m}\right)\sin \left(-\pi t\right)\end{array}$

Conclusion:

Therefore, the equation of motion for the left end of string is $0.100\sin \left(-\pi t\right)$.

(f)

To determine

The point on the string $x=1.50\text{m}$ to the right of left end.

Answer to Problem 16.13P

The equation of motion for the left end of string is $0.100\sin \left(4.71-\pi t\right)$.

Explanation of Solution

Given info: The wavelength of wave is $2.00\text{m}$, the amplitude is $0.100\text{m}$. The speed of wave in string is $1.00\text{m}/\text{s}$. The left end of wave is at origin at $t=0$.

From equation (3),

$y=\left(0.100\text{m}\right)\sin \left(\pi x-\pi t\right)$

For the point $1.50\text{m}$ to the right of the left end of the string:

Substitute $1.50\text{m}$ for $x$ in the above expression.

$y=\left(0.100\text{m}\right)\sin \left(\pi \left(1.50\text{m}\right)-\pi t\right)$

Solve the above expression for $y$,

$y=\left(0.100\text{m}\right)\sin \left(4.71-\pi t\right)$

Conclusion:

Therefore, the equation of motion for the left end of string is $0.100\sin \left(4.71-\pi t\right)$.

(g)

To determine

The maximum speed of element of the string.

Answer to Problem 16.13P

The maximum speed of element of string is $0.314\text{m}/\text{s}$.

Explanation of Solution

Given info: The wavelength of wave is $2.00\text{m}$, the amplitude is $0.100\text{m}$. The speed of wave in string is $1.00\text{m}/\text{s}$. The left end of wave is at origin at $t=0$.

From equation (3), the position of the wave is,

$y=\left(0.100\text{m}\right)\sin \left(\pi x-\pi t\right)$

The change in position with respect to time gives the speed.

Differentiate above equation n with respect to time,

$v=\frac{dy}{dt}$

Here,

$v$ is speed of the element of string.

Substitute $\left(0.100\text{m}\right)\sin \left(\pi x-\pi t\right)$ for $y$ in the above expression.

$v=\frac{d\left(\left(0.100\text{m}\right)\sin \left(\pi x-\pi t\right)\right)}{dt}$

Solve the above expression for $v$,

$\begin{array}{c}v=\frac{d\left(\left(0.100\text{m}\right)\sin \left(\pi x-\pi t\right)\right)}{dt}\\ =\left(0.100\text{m}\right)\frac{d\sin \left(\pi x-\pi t\right)}{dt}+\sin \left(\pi x-\pi t\right)\frac{d\left(0.100\text{m}\right)}{dt}\\ =\left(0.100\text{m}\right)\cos \left(\pi x-\pi t\right)\left(-\pi \right)+0\end{array}$

Substitute $3.14$ for $\pi$ in the above expression.

$\begin{array}{l}=\left(0.100\text{m}\right)\cos \left(\left(3.14\right)x-\left(3.14\right)t\right)\left(-3.14\right)+0\\ =-0.314\text{m}\cos \left(\left(3.14\right)x-\left(3.14\right)t\right)\end{array}$

As the cosine wave varies from the positive of $+1$ to the negative of $+1$. So the maximum speed of any element of string is $0.314\text{m}/\text{s}$.

Conclusion:

Therefore, the maximum speed of element of string is $0.314\text{m}/\text{s}$.