{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

PROBLEM 13 & 14 April 9 2010

PROBLEM 13 & 14 April 9 2010 - SOLVING PROBLEM...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: SOLVING PROBLEM 13 April 12, 2010 (like 14.08): Loudspeakers A and B are located in the x-y plane at plane at yA=0 and yg=l m and x = 0 for both. The speakers are driven at 343 Hz with phases Q, = 0 and #53 = Irradians. What is the smallest value of x >0 on the x-axis at which there is an antinode? Assume the velocity of sound is 343 mfs. a) 7.5 in; b) 0.175 m; c) 1.?5 m; d) 1?.5 m; e) 0.75 m. What do you know? (1) An antinode occurs where the waves arrive in phase. (2) The phase difference from one point to another is two pi times the path length divided by the wavelength it. The wavelength is A = 1 m. The path lengths are x and r = 1le + y; , where y” = it at the chosen frequency. Since the waves begin out of phase, the condition for the first antinode being at the position (Jay) is r - x = mi. 01' 1(it: +342: —x=%}t Squaring we get at2 + y; = 11-212 + ix + J4:2 or since ya = 2. , after eliminating x2, we have x = 3-2 SOLVING PROBLEM 14 (like 14.26): A thin wire with mass per unit length 0.001 kg/m is under a tension of 1 N. The string is stretched along the x-axis where it is fixed (has nodes) at x=0 and x=L= 1m. What is the fundamental vibration frequency (i.e. the lowest resonant frequency)? a) 100 Hz; b) 1000 Hz; 0) 50 Hz; (1) 500 Hz; e) 5000 Hz. What do you know? fl] = VJ?» = vJZL = (l 00 mfs)!'2m = 50 Hz What is the resonant frequency when the string is forced to have a node at the position x=0.2m? a) 0.20 Hz; b) 250 Hz; c) 2500 Hz; d) 1000 Hz; e) 50 Hz. What do you know? The distance between nodes will be US of the previous case, so the frequency will be 5x higher or 250 Hz. ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern