Question

(a) Write an equation for a surface seismic wave moving along the \(-x\) -axis with amplitude \(2.0 \mathrm{cm},\) period \(4.0 \mathrm{s}\) and wavelength $4.0 \mathrm{km} .\( Assume the wave is harmonic, \)x\( is measured in \)\mathrm{m},$ and \(t\) is measured in \(\mathrm{s}\). (b) What is the maximum speed of the ground as the wave moves by? (c) What is the wave speed?

Short answer

Question: Write the equation of a surface seismic wave with a given amplitude of 2.0 cm, period of 4.0 s, and wavelength of 4.0 km, and determine its maximum ground speed and wave speed. Answer: The equation of the surface seismic wave is \(y(x,t)= 0.02\,\text{m}\, \sin\left(\dfrac{2\pi}{4000} (-x) - \dfrac{2\pi}{4} t \right)\). The maximum ground speed is \(0.01\pi\, \text{m/s}\), and the wave speed is \(1000\, \text{m/s}\).

Step-by-step solution

Write the wave equation

First, we need to write the equation for the surface seismic wave. The general form of a harmonic wave equation is: \(y(x,t) = A \sin(kx - \omega t + \phi)\). In this exercise, we are given the amplitude, period, and wavelength, so we can find the wave number (\(k\)) and angular frequency (\(\omega\)) using the following formulas: \(k = \dfrac{2\pi}{\lambda}\) and \(\omega = \dfrac{2\pi}{T}\), where \(\lambda\) is the wavelength and \(T\) is the period. Using the given values, we have: \(A = 2.0\,\text{cm}=0.02\,\text{m}\); \(\lambda = 4.0\, \text{km} = 4000\,\text{m}\); \(T = 4.0\,\text{s}\) Now, let's calculate the wave number (\(k\)) and angular frequency (\(\omega\)): \(k=\dfrac{2\pi}{\lambda}=\dfrac{2\pi}{4000}\, \text{m}^{-1}\) and \(\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{4}\, \text{s}^{-1}\) Assuming the wave is harmonic and the phase constant (\(\phi\)) is \(0\), we can write the wave equation: \(y(x,t)= 0.02\,\text{m}\, \sin\left(\dfrac{2\pi}{4000} x - \dfrac{2\pi}{4} t \right)\) Since the wave is moving along the \(-x\) -axis, we can modify the equation to: \(y(x,t)= 0.02\,\text{m}\, \sin\left(\dfrac{2\pi}{4000} (-x) - \dfrac{2\pi}{4} t \right)\).

Compute the maximum ground speed

The maximum ground speed (i.e., particle speed) can be obtained by computing the derivative of the wave function \(y(x,t)\) with respect to time (\(t\)) and setting \(x=0\): \(V_{max} = \left|\dfrac{\partial y(x,t)}{\partial t}\right|_{x=0}\) We differentiate the wave function with respect to time: \(\dfrac{\partial y(x,t)}{\partial t} = -0.02(\dfrac{2\pi}{4})\,\text{m}\, \cos\left(\dfrac{2\pi}{4000} (-x) - \dfrac{2\pi}{4} t \right)\) Evaluate the expression for the maximum ground speed at \(x=0\): \(V_{max} = \left| -0.02(\dfrac{2\pi}{4})\,\text{m}\, \right| = 0.01\pi\, \text{m/s}\) The maximum ground speed as the wave moves by is \(0.01\pi\, \text{m/s}\).

Determine the wave speed

Finally, to determine the wave speed (\(v\)), we can use the equation: \(v = \dfrac{\lambda}{T}\) Using the given values, we have: \(v = \dfrac{4000\,\text{m}}{4\,\text{s}}= 1000\,\text{m/s}\) The wave speed is \(1000\, \text{m/s}\).