{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Hmwk_12_Solutions

# Hmwk_12_Solutions - Physics 221 Fall 2008 Homework#12...

This preview shows pages 1–2. Sign up to view the full content.

1 Physics 221 Fall 2008 Homework #12 Solutions Ch. 16 Due Tues, Nov 18, 2008 12.1 A longitudinal sinusoidal sound wave in monatomic neon (Ne) gas at atmospheric pressure has a frequency of 440 Hz and a wave speed of 380 m/s. The wave is moving in the – x direction. The amplitude of the displacement wave associated with the sound wave is 4.00 nm. The molar mass of Ne is 0.0202 kg/mol and the ratio of heat capacities is ± = 1.67. (a) Explain why a pressure wave is also associated with the sound wave. What is the amplitude of the pressure wave? At a peak in the displacement wave, a slab of air has moved to the right. At the trough to the right of the peak, a slab of air has moved to the left. Thus in between the crest and the trough, the gas is compressed, resulting in an increase in pressure above ambient. To the left of the crest, the pressure is lower than ambient. This pressure wave moves along with the displacement wave but is shifted by 1/4 of a wavelength from it. B = ± p 0 = (1.67)(1.01 ² 10 5 Pa) = 1.69 ² 10 5 Pa . ³ = 2 ´ f = 2 ´ (440 Hz) = 2760 rad/s k = ³ / v = (2760 rad/s)/(380 m/s) = 7.26 rad/m . A = 4.00 nm . The amplitude of the pressure wave is p max = BAk = (1.69 ² 10 5 Pa)(4.00 ² 10 µ 9 m)(7.26 rad/m) = 4.91 mPa . (b) Write down a possible wave function for the displacement wave. Then derive from this using Eq. (16.3) the wave function for the pressure wave. How are these two wave functions related to each other? Let y ( x , t ) = A sin( kx + ³ t ) . The “+” sign: the wave is moving towards the – x direction. p ( x , t ) = µ B y ( x , t )/ x = µ B [ A sin( kx + ³ t )]/ x = µ BAk cos( kx + ³ t ) .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern