Chapter 16, Problem 16.15P

A transverse wave on a siring is described by the wave function

$y=0.120\sin \left(\frac{\pi }{8}x+4\pi t\right)$

where x and y are in meters and t is in seconds. Determine (a) the transverse speed and (b) the transverse acceleration at t = 0.200 s for an element of the string located at x = 1.60 m. What are (c) the wavelength, (d) the period, and (e) the speed of propagation of this wave?

To determine

The transverse speed for an element located at $1.60\text{m}$.

Answer to Problem 16.15P

The transverse speed for an element located at $1.60\text{m}$ is $-1.51\text{m}/\text{s}$.

Explanation of Solution

Given info: The wave function of the wave is $y=0.120\text{sin}\left(\frac{\pi }{8}x+4\pi t\right)$.

The standard equation of the transverse wave is,

$y=A\sin \left(kx-\omega t\right)$ . (1)

Here,

$A$ is amplitude of wave.

$k$ is angular wave number of wave.

$\omega$ is angular velocity of wave.

The wave function give is,

$y=0.120\text{sin}\left(\frac{\pi }{8}x+4\pi t\right)$ . (2)

Compare the equation (1) and equation (2).

$\begin{array}{c}A=0.120\text{m}\\ k=\frac{\pi }{8}\\ \omega =4\pi \text{rad}/\text{s}\end{array}$

The formula to calculate frequency is,

$f=\frac{\omega }{2\pi }$

Substitute $4\pi \text{rad}/\text{s}$ for $\omega$ in the above expression.

$\begin{array}{c}f=\frac{4\pi \text{rad}/\text{s}}{2\pi }\\ =2.00\text{Hz}\end{array}$

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

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

Here,

$dy$ is change in position.

$dt$ is change in time.

Substitute $0.120\text{sin}\left(\frac{\pi }{8}x+4\pi t\right)$ for $y$ in the above expression.

$v=\frac{d\left(0.120\text{sin}\left(\frac{\pi }{8}x+4\pi t\right)\right)}{dt}$

Differentiate and solve the above expression for $v$,

$\begin{array}{l}=\frac{\left(0.120\right)d\text{sin}\left(\frac{\pi }{8}x+4\pi t\right)+\text{sin}\left(\frac{\pi }{8}x+4\pi t\right)\frac{d\left(0.120\right)}{dt}}{dt}\\ =\left(0.120\right)\cos \left(\frac{\pi }{8}x+4\pi t\right)\left(4\pi \right)+0\end{array}$ (1)

Substitute $0.200\text{s}$ for $t$ and $1.6\text{m}$ for $x$ in the above expression.

$\begin{array}{c}=\left(0.120\right)\cos \left(\frac{\pi }{8}\left(1.6\text{m}\right)+4\pi \left(0.200\text{s}\right)\right)\left(4\pi \right)+0\\ =\left(0.120\right)\left(4\pi \right)\cos \left(0.2\pi +0.8\pi \right)\\ =-1.5072\text{m}/\text{s}\\ \approx -1.51\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the transverse speed for an element located at $1.60\text{m}$ is $-1.51\text{m}/\text{s}$.

To determine

To write: The transverse acceleration of the wave.

Answer to Problem 16.15P

The transverse acceleration of the wave is $0$.

Explanation of Solution

Given info: The wave function of the wave is $y=0.120\text{sin}\left(\frac{\pi }{8}x+4\pi t\right)$. The time of the wave is $0.200\text{s}$ and the position of element is $1.60\text{m}$.

The change in velocity with respect to time gives the acceleration.

$a=\frac{dv}{dt}$ (2)

Here,

$dv$ is change in velocity.

$dt$ is change in time.

From equation (1), the speed is,

$v=\left(0.120\right)\cos \left(\frac{\pi }{8}x+4\pi t\right)\left(4\pi \right)+0$

Substitute $\left(0.120\right)\cos \left(\frac{\pi }{8}x+4\pi t\right)\left(4\pi \right)+0$ for $v$ in equation (2).

$a=\frac{d\left(0.120\right)\cos \left(\frac{\pi }{8}x+4\pi t\right)\left(4\pi \right)+0}{dt}$

Differentiate and solve the above expression for $a$.

$\begin{array}{l}a=\frac{d\left(0.120\right)\cos \left(\frac{\pi }{8}x+4\pi t\right)\left(4\pi \right)+0}{dt}\\ =\left(0.120\right)\times \left(-\sin \left(\left(\frac{\pi }{8}x+4\pi t\right)\right)\times {\left(4\pi \right)}^2\right)+0\end{array}$

Substitute $0.200\text{s}$ for $t$ and $1.60\text{m}$ for $x$ in the above expression.

$\left(0.120\right)\times \left(-\sin \left(\left(\frac{\pi }{8}\left(1.60\text{m}\right)+4\pi \left(0.200\text{s}\right)\right)\right)\times {\left(4\pi \right)}^2\right)+0$

Solve the above expression.

$\begin{array}{c}=\left(0.120\right)\times \left(-\sin \left(\left(\frac{\pi }{8}\left(1.60\text{m}\right)+4\pi \left(0.200\text{s}\right)\right)\right)\times {\left(4\pi \right)}^2\right)+0\\ =-\left(0.120\right){\left(4\pi \right)}^2\left(-\sin \left(\pi \right)\right)\\ =0\end{array}$

Conclusion:

Therefore, the transverse acceleration of the wave is $0$.

To determine

The wavelength of the wave.

Answer to Problem 16.15P

The wavelength of the wave is $16.0\text{m}$.

Explanation of Solution

Given info: The wave function of the wave is $y=0.120\text{sin}\left(\frac{\pi }{8}x+4\pi t\right)$.

The formula to calculate wavelength of the wave is,

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

Here,

$k$ is angular wave number of wave.

Substitute $\frac{\pi }{8}$ for $k$ in the above expression.

$\lambda =\frac{2\pi }{\left(\frac{\pi }{8}\right)}$

Solve the above expression for $\lambda$.

$\begin{array}{c}\lambda =\frac{2\pi }{\left(\frac{\pi }{8}\right)}\\ =2\times 8\text{m}\\ =16.0\text{m}\end{array}$

Conclusion:

Therefore, the wavelength of the wave is $16.0\text{m}$.

To determine

The period of the wave.

Answer to Problem 16.15P

The period of the wave is $0.500\text{s}$.

Explanation of Solution

Given info: The wave function of the wave is $y=0.120\text{sin}\left(\frac{\pi }{8}x+4\pi t\right)$.

The formula to calculate frequency is,

$f=\frac{\omega }{2\pi }$

Here,

$\omega$ is angular frequency of wave.

Substitute $4\pi \text{rad}/\text{s}$ for $\omega$ in the above expression.

$\begin{array}{c}f=\frac{4\pi \text{rad}/\text{s}}{2\pi }\\ =2\text{Hz}\end{array}$

The formula to calculate time period of the wave is,

$t=\frac{1}{f}$

Here,

$f$ is frequency of wave.

Substitute $2\text{Hz}$ for $f$ in the above expression.

$\begin{array}{c}t=\frac{1}{2\text{Hz}}\\ =0.500\text{s}\end{array}$

Conclusion:

Therefore, the period of the wave is $0.500\text{s}$.

To determine

The speed of propagation of wave.

Answer to Problem 16.15P

The speed of propagation of wave is $32.0\text{m}/\text{s}$.

Explanation of Solution

Given info: The wave function of the wave is $y=0.120\text{sin}\left(\frac{\pi }{8}x+4\pi t\right)$.

The formula to calculate speed of propagation of wave is,

$s=f\lambda$

Here,

$s$ is speed of propagation of wave.

$f$ is frequency of wave.

Substitute $2{\text{s}}^{-1}$ for $f$ and $16.0\text{m}$ for $\lambda$ in the above expression.

$\begin{array}{c}s=\left(2{\text{s}}^{-1}\right)\left(16.0\text{m}\right)\\ =32.0\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed of propagation of wave is $32.0\text{m}/\text{s}$.