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PROBLEM 13 &amp; 14 April 9 2010

# PROBLEM 13 &amp; 14 April 9 2010 - SOLVING PROBLEM...

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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. ...
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