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Chapter 15: Problem 25

A sinusoidal wave traveling in the positive \(x\) -direction has a wavelength of \(12 \mathrm{~cm},\) a frequency of \(10.0 \mathrm{~Hz},\) and an amplitude of \(10.0 \mathrm{~cm}\). The part of the wave that is at the origin at \(t=0\) has a vertical displacement of \(5.00 \mathrm{~cm} .\) For this wave, determine the a) wave number, d) speed, b) period, e) phase angle, and c) angular frequency, f) equation of motion.

Short Answer

Expert verified
Based on the given information and step-by-step solutions, we found the following properties of the sinusoidal wave: 1. Wave number (k) = 52.36 m⁻¹ 2. Period (T) = 0.1 s 3. Angular frequency (ω) = 20π rad/s 4. Speed (v) = 1.2 m/s 5. Phase angle (φ) = 30° Finally, the equation of motion for the given wave is: y(x, t) = 10.0 cm sin(52.36 m⁻¹x - (20π rad/s)t + 30°)

Step by step solution

01

Wave number

The wave number (denoted by \(k\)) is the spatial frequency of the wave. To find the wave number, we can use the following equation: $$k = \frac{2\pi}{\lambda}$$ Here, \(\lambda\) is the wavelength which is given as \(12 cm\). So, let's calculate the wave number: $$k = \frac{2\pi}{12\ \text{cm}} = \frac{2\pi}{0.12\ \text{m}} \approx 52.36\ \text{m}^{-1}$$
02

Period

The period (denoted by \(T\)) is the time taken for one complete oscillation. It is related to the frequency (denoted by \(f\)) as follows: $$T = \frac{1}{f}$$ Here, the frequency is given as \(10.0 Hz\). So, let's calculate the period: $$T = \frac{1}{10.0\ \text{Hz}} = 0.1\ \text{s}$$
03

Angular frequency

The angular frequency (denoted by \(\omega\)) is related to the frequency by the following equation: $$\omega = 2\pi f$$ We have the frequency (\(10.0 Hz\)). Let's calculate the angular frequency: $$\omega = 2\pi (10.0\ \text{Hz}) = 20\pi\ \text{rad/s}$$
04

Speed

The wave speed (denoted by \(v\)) is the speed at which the wave travels in the medium. It can be found using the following equation: $$v = \lambda f$$ We have the wavelength (\(12 cm\)) and the frequency (\(10.0 Hz\)). Let's calculate the wave speed: $$v = (12\ \text{cm})(10.0\ \text{Hz}) = 1.2\ \text{m/s}$$
05

Phase angle

To find the phase angle (denoted by \(\phi\)), we need to consider the vertical displacement at the origin when \(t=0\). Let's denote the vertical displacement as \(y\). The wave equation is given by: $$y(x, t) = A\sin(kx - \omega t + \phi)$$ At \(t=0\) and \(x=0\), \(y(0,0) = 5.00\ \text{cm}\). Let's substitute the values and solve for \(\phi\): $$5.00\ \text{cm} = (10.0\ \text{cm})\sin(\phi)$$ $$\phi = \arcsin\left(\frac{1}{2}\right)$$ $$\phi = 30^{\circ}$$
06

Equation of motion

Now we have all the necessary parameters to write the equation of motion for the given wave. The wave equation is given by: $$y(x, t) = A\sin(kx - \omega t + \phi)$$ Substitute the values we found for \(A\), \(k\), \(\omega\), and \(\phi\): $$y(x, t) = 10.0\ \text{cm}\sin\left(52.36\ \text{m}^{-1}x - (20\pi\ \text{rad/s})t + 30^{\circ}\right)$$

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Most popular questions from this chapter

The middle-C key (key 52 ) on a piano corresponds to a fundamental frequency of about \(262 \mathrm{~Hz},\) and the sopranoC key (key 64) corresponds to a fundamental frequency of \(1046.5 \mathrm{~Hz}\). If the strings used for both keys are identical in density and length, determine the ratio of the tensions in the two strings.

A traveling wave propagating on a string is described by the following equation: $$ y(x, t)=(5.00 \mathrm{~mm}) \sin \left(\left(157.08 \mathrm{~m}^{-1}\right) x-\left(314.16 \mathrm{~s}^{-1}\right) t+0.7854\right) $$ a) Determine the minimum separation, \(\Delta x_{\min }\), between two points on the string that oscillate in perfect opposition of phases (move in opposite directions at all times). b) Determine the separation, \(\Delta x_{A B}\), between two points \(A\) and \(B\) on the string, if point \(B\) oscillates with a phase difference of 0.7854 rad compared to point \(A\). c) Find the number of crests of the wave that pass through point \(A\) in a time interval \(\Delta t=10.0 \mathrm{~s}\) and the number of troughs that pass through point \(B\) in the same interval. d) At what point along its trajectory should a linear driver connected to one end of the string at \(x=0\) start its oscillation to generate this sinusoidal traveling wave on the string?

Suppose that the tension is doubled for a string on which a standing wave is propagated. How will the velocity of the standing wave change? a) It will double. c) It will be multiplied by \(\sqrt{2}\). b) It will quadruple. d) It will be multiplied by \(\frac{1}{2}\).

A sinusoidal wave on a string is described by the equation \(y=(0.100 \mathrm{~m}) \sin (0.75 x-40 t),\) where \(x\) and \(y\) are in meters and \(t\) is in seconds. If the linear mass density of the string is \(10 \mathrm{~g} / \mathrm{m}\), determine (a) the phase constant, (b) the phase of the wave at \(x=2.00 \mathrm{~cm}\) and \(t=0.100 \mathrm{~s}\) (c) the speed of the wave, (d) the wavelength, (e) the frequency, and (f) the power transmitted by the wave.

A string is made to oscillate, and a standing wave with three antinodes is created. If the tension in the string is increased by a factor of 4 a) the number of antinodes increases. b) the number of antinodes remains the same. c) the number of antinodes decreases. d) the number of antinodes will equal the number of nodes.
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