Chapter 11, Problem 82P

(a)

To determine

An equation for a surface seismic wave.

Answer to Problem 82P

An equation for a surface seismic wave is $y\left(x,t\right)=\left(0.020\text{m}\right)\sin \left(\left(1.6\text{rad/s}\right)t+\left(0.0016\text{rad/m}\right)x\right)$.

Explanation of Solution

The wave is moving is $-x$ axis. The amplitude of the wave is $2.0\text{cm}$. The period of the wave is $4.0\text{s}$. The wave length of the wave is $4.0\text{km}$. The wave is considered to be harmonic.

Write the formula for the equation of a harmonic wave.

$y\left(x,t\right)=A\sin \left(\omega t+kx\right)$ (I)

Here, $y$ is the displacement, $A$ is the amplitude, $\omega$ is the angular speed, $t$ is the time, $k$ is the propagation vector, $x$ is the distance travelled by the wave.

Write the formula for $\omega$ in terms of time period.

$\omega =\frac{2\pi }{T}$ (II)

Here, $T$ is the time period.

Write the formula for $k$ in terms of wavelength.

$k=\frac{2\pi }{\lambda }$ (III)

Here, $\lambda$ is the wavelength.

Substitute equation (II) and (III) in equation (I).

$y\left(x,t\right)=A\sin \left(\frac{2\pi t}{T}+\frac{2\pi x}{\lambda }\right)$ (IV)

Conclusion:

Substitute $2.0\text{cm}$ for $A$, $4.0\text{s}$ for $T$, $4.0\text{km}$ for $\lambda$ in equation (IV).

$\begin{array}{c}y\left(x,t\right)=\left(2.0\text{cm}\right)\sin \left(\frac{2\pi t}{4.0\text{s}}+\frac{2\pi x}{4.0\text{km}}\right)\\ =\left(20\times {10}^{-2}\text{m}\right)\sin \left(\frac{2\pi t}{4.0\text{s}}+\frac{2\pi x}{4.0\times {10}^3\text{m}}\right)\\ =\left(0.020\text{m}\right)\sin \left(\left(1.6\text{rad/s}\right)t+\left(0.0016\text{rad/m}\right)x\right)\end{array}$

An equation for a surface seismic wave is $y\left(x,t\right)=\left(0.020\text{m}\right)\sin \left(\left(1.6\text{rad/s}\right)t+\left(0.0016\text{rad/m}\right)x\right)$.

(b)

To determine

The maximum speed of ground as the wave pass by it.

Answer to Problem 82P

The maximum speed of ground as the wave pass by it is $0.031\text{m/s}$.

Explanation of Solution

The wave is moving is $-x$ axis. The amplitude of the wave is $2.0\text{cm}$. The period of the wave is $4.0\text{s}$. The wave length of the wave is $4.0\text{km}$. The wave is considered to be harmonic.

Write the formula for the maximum speed of the ground as the wave pass by.

$\begin{array}{c}{v}_m=\omega A\\ =\frac{2\pi A}{T}\end{array}$ (V)

Here, $v_m$ is the maximum speed, $A$ is the amplitude, $\omega$ is the angular speed, $T$ is the time period.

Conclusion:

Substitute $2.0\text{cm}$ for $A$, $4.0\text{s}$ for $T$ in equation (V).

$\begin{array}{c}{v}_m=\frac{2\pi \left(2.0\text{cm}\right)}{4.0\text{s}}\\ =\frac{2\pi \left(2.0\times {10}^{-2}\text{m}\right)}{4.0\text{s}}\\ =0.031\text{m/s}\end{array}$

The maximum speed of ground as the wave pass by it is $0.031\text{m/s}$.

(c)

To determine

The speed of the wave.

Answer to Problem 82P

The speed of the wave is $1.0\text{km/s}$.

Explanation of Solution

The wave is moving is $-x$ axis. The amplitude of the wave is $2.0\text{cm}$. The period of the wave is $4.0\text{s}$. The wave length of the wave is $4.0\text{km}$. The wave is considered to be harmonic.

Write the formula for the speed of the propagation of the wave.

$\begin{array}{c}v=\lambda f\\ =\frac{\lambda }{T}\end{array}$ (VI)

Here, $v$ is the propagation speed of the wave, $\lambda$ is the wavelength, $f$ is the frequency, $T$ is the time period.

Conclusion:

Substitute $4.0\text{km}$ for $\lambda$, $4.0\text{s}$ for $T$ in equation (VI).

$\begin{array}{c}v=\frac{4.0\text{km}}{4.0\text{s}}\\ =1.0\text{km/s}\end{array}$

The speed of the wave is $1.0\text{km/s}$.