hw37 - This print-out should have 17 questions...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 10.0 points A traveling wave propagates according to the expression y = (5 . 33 cm) sin bracketleftBig (1 . 93 cm − 1 ) x- (3 . 58 s − 1 ) t bracketrightBig , where x is in centimeters, and t is in seconds. Find the amplitude of the wave. Correct answer: 5 . 33 cm. Explanation: Let : A = 5 . 33 cm . Given a wave y = A sin( k x- ω t ), the ampli- tude of the wave is A = 5 . 33 cm . 002 (part 2 of 4) 10.0 points Determine the wavelength of the wave. Correct answer: 3 . 25554 cm. Explanation: Let : k = 1 . 93 cm − 1 . The angular wave number of the wave is k , so λ = 2 π k = 2 π 1 . 93 cm − 1 = 3 . 25554 cm . 003 (part 3 of 4) 10.0 points Determine the frequency of the wave. Correct answer: 0 . 569775 Hz. Explanation: Let : ω = 3 . 58 s − 1 . The angular frequency of the wave is ω , so f = ω 2 π = 3 . 58 s − 1 2 π = . 569775 Hz . 004 (part 4 of 4) 10.0 points Determine the period of the wave. Correct answer: 1 . 75508 s. Explanation: The period of the wave is T = 1 f = 1 . 569775 Hz = 1 . 75508 s . 005 10.0 points The red light emitted by a He-Ne laser has a wavelength of 633 nm in air and travels at 3 . 00 × 10 8 m/s. Find the frequency of the laser light. Correct answer: 4 . 73934 × 10 14 Hz. Explanation: Let : λ = 633 nm and v = 3 × 10 8 m / s . The speed of light is given by v = f λ f = v λ = 3 × 10 8 m / s 633 nm · 10 9 nm 1 m = 4 . 73934 × 10 14 Hz . 006 10.0 points A wave on a string is described by the wave function y = (0 . 1 m) sin[(0 . 5 rad / m) x- (20 rad / s) t ] Determine the frequency of oscillation of a particular point at x = 2 . 0 m. Correct answer: 3 . 1831 Hz. Explanation: Let : ω = 20 rad / s . In fact, when a wave with frequency f trav- els along a string, any point of the string has the same oscillation frequency f . In this case, f = ω 2 π = 20 rad / s 2 π = 3 . 1831 Hz keywords: 007 (part 1 of 2) 10.0 points A steel piano wire is 0 . 7 m long and has a homework 37 – Turner – (58185) 1 hinojosa (jlh3938) – homework 37 – Turner – (58185)...
View Full Document

This note was uploaded on 05/02/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

Page1 / 5

hw37 - This print-out should have 17 questions...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online