The two speakers of a boom box are 37.0 cm apart. A single oscillator makes the
ID: 1730458 • Letter: T
Question
The two speakers of a boom box are 37.0 cm apart. A single oscillator makes the speakers vibrate in phase at a frequency of 2.01 kHz. (Use 340 m/s as the speed of sound.) At what nonnegative angles, measured from the perpendicular bisector of the line joining the speakers, would a distant observer hear maximum sound intensity? Enter all of these angles using hundredths precision in the single answe box below and separate your results with spaces. For example, 0.00,9.32, and 33.74 could be entered as 0.00 33.74 9.32. At what nonnegative angles would such an observer hear minimum sound intensity? Enter all of these angles using hundredths precision in the single answer box below and separate your results with spaces. For example, 0.00°, 9.32° and 33.74 could be entered as 0.00 33.74 9.32.Explanation / Answer
Given,
Distance, d = 37 cm = 0.37 m
Frequency, f = 2.01 kHz = 2010 Hz
Speed of sound, v = 340 m / s
As we know that, v = f ?
wavelength, ? = v / f = (340 m/s) / (2010 Hz) = 0.169 m
we have, d sin ? = m ?
so the angle for maximal ? = sin-1[m(? / d)]
(? / d) = 0.457
where m = 0, 1, 2, ...
For m = 0, ? = 0°
? = sin-1(0.457) = 27.20°
the angle for minima, ? = sin-1[(2 m + 1) (? / 2d)]
solve for m = 0 and 1
? = 13.20°
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