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Two identical loudspeakers are driven in phase by a common oscillator at 800 Hz and face each other at a distance of 1.25 m. Locate the points along the line joining the two speakers where relative minimum of sound pressure amplitude would be expected. ( Use v = 343 m//s.) |
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Answer» Solution :TWO identical WAVES moving in opposite directions constitute a standing WAVE .We must find the nodes. The wavelengthis ` lambda = (v)/(f) = ( 343 m//s)/( 800 Hz) = 0.429 m` The two waves moving in opposite directions along the line between the two speakers will add to produce a standing wave within this distance between nodes. Distance node to node ` = lambda//2 = 0.214 m` Because the speakers vibrate in phase , air compressions from each will simultaneously reach the POINT halfway between the speakers, to produce antinode of pressure here . A node of pressure will be located at this distance on either side of the midpoint. Distance node to antinode ` = lambda //4 = 0.107 m` Therefore nodes of sound pressure will appear at these distances from either speaker. `[ 1//2 ( 1.25 m) + 0.107 m = 0.732 m` and `1//2 ( 1.25 m) - 0.107 m = 0.518 m ]` The standing wave CONTAINS a chain of equally spaced nodes at distances from either of the speakers `0.732 m + 0.214 m = 0.946 m` ` 0.947 m + 0.214 m = 1.16 m` also at ` 0.518 m - 0.214 m = 0.303 m` `0.303 m - 0.214 m = 0.089 m` The standing wave exists only along the line segment between the speakers . No nodes or antinodes appear at distances greater than `1.252 m` or less than `0`, because waves add to gives standing wave only if they are travelling in opposite directions and not in the same direction . Thus , the distances from either of the speakers to the nodes of pressure between the speakers are `0.089 m , 0.303 m , 0.732 m , 0.946 m and 1.16 m` |
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