Answer Key Chapter 16 - University Physics Volume 1 _ OpenStax (2).pdf

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8/7/2020Answer Key Chapter 16 - University Physics Volume 1 | OpenStax1/7Check Your Understanding16.1The wavelength of the waves depends on the frequency and the velocity of the wave. The frequencyof the sound wave is equal to the frequency of the wave on the string. The wavelengths of the sound wavesand the waves on the string are equal only if the velocities of the waves are the same, which is not alwaysthe case. If the speed of the sound wave is different from the speed of the wave on the string, thewavelengths are different. This velocity of sound waves will be discussed inSound.16.2In a transverse wave, the wave may move at a constant propagation velocity through the medium, butthe medium oscillates perpendicular to the motion of the wave. If the wave moves in the positivex-direction, the medium oscillates up and down in they-direction. The velocity of the medium is therefore notconstant, but the medium’s velocity and acceleration are similar to that of the simple harmonic motion of amass on a spring.16.3Yes, a cosine function is equal to a sine function with a phase shift, and either function can be used ina wave function. Which function is more convenient to use depends on the initial conditions. InFigure16.11, the wave has an initial height ofy(0.00, 0.00) = 0and then the wave height increases to the maximumheight at the crest. If the initial height at the initial time was equal to the amplitude of the wavey(0.00, 0.00) = +A,then it might be more convenient to model the wave with a cosine function.16.4This wave, with amplitudeA= 0.5 m,wavelengthλ= 10.00 m,periodT= 0.50 s,is a solution to thewave equation with a wave velocityv= 20.00 m/s.16.5Since the speed of a wave on a taunt string is proportional to the square root of the tension divided bythe linear density, the wave speed would increase by2.16.6At first glance, the time-averaged power of a sinusoidal wave on a string may look proportional to thelinear density of the string becauseP=12μA2ω2v;however, the speed of the wave depends on the lineardensity. Replacing the wave speed withFTμshows that the power is proportional to the square root oftension and proportional to the square root of the linear mass density:P=12μA2ω2v=12μA2ω2FTμ=12A2ω2μFT.16.7Yes, the equations would work equally well for symmetric boundary conditions of a medium free tooscillate on each end where there was an antinode on each end. The normal modes of the first three modesare shown below. The dotted line shows the equilibrium position of the medium.

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