Homework 20 Solutions

# Homework 20 Solutions - lim (kl9356) – Homework 20 –...

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

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.

Unformatted text preview: lim (kl9356) – Homework 20 – Weathers – (17104) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points It is found that a 6 . 1 m segment of a long string contains 3 complete wavelengths and has a mass of 190 g. At one point it is vibrat- ing sinusoidally with a frequency of 55 Hz and a peak to valley displacement of 0 . 8 m. What power is being transmitted along the string? Correct answer: 33279 W. Explanation: Let : l = 6 . 1 m , λ = 6 . 1 m 3 , m = 190 g , f = 55 Hz , and A = . 8 m 2 . Since k ≡ 2 π λ , ω ≡ 2 π f , v ≡ ω k , P = 1 2 μ ω 2 A 2 v = 1 2 parenleftBig m l parenrightBig (2 π f ) 2 A 2 λ f = 1 2 parenleftbigg 190 g 6 . 1 m parenrightbigg [2 π (55 Hz)] 2 × parenleftbigg . 8 m 2 parenrightbigg 2 parenleftbigg 6 . 1 m 3 parenrightbigg (55 Hz) = 33279 W . 002 (part 1 of 2) 10.0 points A point source emits sound waves with an average power output of 1 . 24 W. Find the intensity 3 . 15 m from the source. Correct answer: 0 . 00994468 W / m 2 . Explanation: A point source emits energy in the form of a spherical waves. At a distance R from the source, the power is distributed over the surface area of a sphere, 4 πR 2 . Therefore, the intensity at a distance R = 3 . 15 m from the source is I = P av 4 π R 2 = 0 . 00994468 W / m 2 . 003 (part 2 of 2) 10.0 points Find the distance at which the sound reduces to a level of 64 dB. Correct answer: 198 . 201 m. Explanation: We can find the intensity at the β = 64 dB level by using the equation β = 10 log parenleftbigg I I parenrightbigg with I = 1 × 10 − 12 W / m 2 : 64 dB = 10 log parenleftbigg I 1 × 10 − 12 W / m 2 parenrightbigg therefore I = 10 β/ 10 I = 10 6 . 4 (1 × 10 − 12 W / m 2 ) = 2 . 51189 × 10 − 6 W / m 2 . Using this value for I and the equation I = P av 4 π r 2 , we find for the distance r = radicalbigg P av 4 π I = radicalBigg 1 . 24 W 4 π (2 . 51189 × 10 − 6 W / m 2 ) = 198 . 201 m ....
View Full Document

## This note was uploaded on 09/07/2010 for the course PHYS 1710 taught by Professor Weathers during the Fall '08 term at North Texas.

### Page1 / 5

Homework 20 Solutions - lim (kl9356) – Homework 20 –...

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

View Full Document
Ask a homework question - tutors are online