{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# oldHWK33 - oldhomewk 33 VALENCIA DANIEL Due 4:00 am...

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

oldhomewk 33 – VALENCIA, DANIEL – Due: Apr 28 2008, 4:00 am 1 Question 1, chap 16, sect 1. part 1 of 1 10 points When a particular wire is vibrating with a frequency of 3 . 1 Hz, a transverse wave of wavelength 38 . 8 cm is produced. Determine the speed of wave pulses along the wire. Correct answer: 1 . 2028 m / s (tolerance ± 1 %). Explanation: If the frequency is f and the wavelength is λ then the speed of wave is v = f λ = (3 . 1 Hz) (38 . 8 cm) (0 . 01 m / cm) = 1 . 2028 m / s Question 2, chap 16, sect 2. part 1 of 1 10 points Radio waves travel at the speed of light: 300000 km / s. What is the wavelength of radio waves re- ceived at 103 . 1 MHz on your FM radio dial? Correct answer: 2 . 9098 m (tolerance ± 1 %). Explanation: Let : v = 300000 km / s = 300 million m / s and f = 103 . 1 MHz = 103 . 1 million Hz . λ = v f = 300 million m / s 103 . 1 million Hz = 2 . 9098 m . Question 3, chap 16, sect 2. part 1 of 1 10 points Two harmonic waves are described by y 1 = A sin( k x - ω t - φ 1 ) and y 2 = A sin( k x - ω t - φ 2 ) , where k = 4 m - 1 , A = 5 m , ω = 504 rad / s , φ 1 = 0 rad , and φ 2 = 4 rad . What is the frequency of the resultant wave y = y 1 + y 2 ? Correct answer: 80 . 2141 Hz (tolerance ± 1 %). Explanation: From the basic trigonometric relation sin θ 1 + sin θ 2 = sin θ 1 + θ 2 2 cos θ 1 - θ 2 2 , where θ 1 = k x - ω t - φ 1 and θ 2 = k x - ω t - φ 2 .

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