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Chapter 11: Problem 70

A seismic wave is described by the equation $y(x, t)=(7.00 \mathrm{cm}) \cos [(6.00 \pi \mathrm{rad} / \mathrm{cm}) x+(20.0 \pi \mathrm{rad} / \mathrm{s}) t]\( The wave travels through a uniform medium in the \)x$ -direction. (a) Is this wave moving right ( \(+x\) -direction) or left (-x-direction)? (b) How far from their equilibrium positions do the particles in the medium move? (c) What is the frequency of this wave? (d) What is the wavelength of this wave? (e) What is the wave speed? (f) Describe the motion of a particle that is at \(y=7.00 \mathrm{cm}\) and \(x=0\) when \(t=0 .\) (g) Is this wave transverse or longitudinal?

Short Answer

Expert verified
In summary, the given wave moves in the negative x-direction (left), with a maximum displacement of 7.00 cm, frequency of 10.0 Hz, wavelength of \(\frac{1}{3.00}\, \mathrm{cm}\), and wave speed of \(\frac{10.0}{3.00}\, \mathrm{cm/s}\). The given particle at \(y=7.00\mathrm{cm}\) and \(x=0\) when \(t=0\) oscillates vertically up and down as the wave moves left. The wave is a transverse wave since the particles oscillate perpendicular to the direction of wave motion.

Step by step solution

01

Write the wave equation in standard form

The given wave equation is \(y(x, t) = (7.00 \mathrm{cm}) \cos([(6.00\pi\mathrm{rad}/\mathrm{cm})x + (20.0\pi\mathrm{rad}/\mathrm{s})t])\). We can represent this equation in standard form as \(y(x, t) = A \cos(kx \pm \omega t)\), where \(A\) is the amplitude, \(k\) is the wave number, and \(\omega\) is the angular frequency.
02

Analyze the wave direction

The wave equation is given in the form \(y(x, t) = A \cos(kx + \omega t)\). Since the sign between \(kx\) and \(\omega t\) is positive, the wave is moving in the negative x-direction. Answer (a): The wave is moving left or in the -\(x\) direction.
03

Find the amplitude

The amplitude of the wave is the maximum displacement of the particles from their equilibrium positions. In the given equation, the amplitude is 7.00 cm. Answer (b): The particles in the medium move 7.00 cm from their equilibrium positions.
04

Find the frequency

The angular frequency \(\omega\) is given as 20.0\(\pi \mathrm{rad} / \mathrm{s}\). To find the frequency (\(f\)) of the wave, we can use the formula \(\omega = 2\pi f\): \(f = \frac{\omega}{2\pi} = \frac{20.0\pi \mathrm{rad} / \mathrm{s}}{2\pi} = 10.0\, \mathrm{Hz}\) Answer (c): The frequency of the wave is 10.0 Hz.
05

Find the wavelength

The wave number \(k\) is given as \((6.00 \pi \mathrm{rad} / \mathrm{cm})\). Using the formula \(k = \frac{2\pi}{\lambda}\) we can find the wavelength \(\lambda\): \(\lambda = \frac{2\pi}{k} = \frac{2\pi}{6.00\pi\mathrm{rad} / \mathrm{cm}} = \frac{1}{3.00}\, \mathrm{cm}\) Answer (d): The wavelength of the wave is \(\frac{1}{3.00}\, \mathrm{cm}\).
06

Find the wave speed

We can find the wave speed (\(v\)) using the formula \(v=\lambda f\): \(v = \lambda f = (\frac{1}{3.00}\, \mathrm{cm})(10.0\, \mathrm{Hz}) = \frac{10.0}{3.00}\, \mathrm{cm/s}\) Answer (e): The wave speed is \(\frac{10.0}{3.00}\, \mathrm{cm/s}\).
07

Describe the motion of the given particle

The problem states that a particle is at \(y=7.00 \mathrm{cm}\) and \(x=0\) when \(t=0\). We can find the particle's position \(y(x, t)\) at different times: At \(t=0\), the wave equation becomes \(y(0, 0) = 7.00 \mathrm{cm} \cos(0) = 7.00 \mathrm{cm}\). Since the particle is at its maximum displacement at \(t=0\), we know that it will move towards the equilibrium position and then back to the maximum displacement in the opposite direction. As the wave moves to the left, the particle will oscillate in the \(y\)-direction. Answer (f): The motion of the particle at \(y=7.00\mathrm{cm}\) and \(x=0\) when \(t=0\) is vertical up and down oscillation as the wave moves to the left.
08

Identify the type of wave

Since the particle moves perpendicular to the direction of wave motion, which is up and down in the \(y\)-direction as the wave moves horizontally in the \(x\)-direction, this wave is a transverse wave. Answer (g): This wave is transverse.

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Most popular questions from this chapter

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