HW5Sol - Oregon State University Physics 202 HW5 Solutions...

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Oregon State University Physics 202 , Winter 2010 HW5 W (due Feb. 8 at 5:00 p.m.) Page 1 Oregon State University Physics 202 Winter Term, 2010 HW5 Solutions 1. Two loudspeakers on a concert stage are vibrating in phase. You are sitting at a distance of 5.50 m from the left speaker and 3.86 m from the right speaker. The speakers are playing music that varies in frequency (i.e. many different frequencies are included in the sound that arrives at your ears), but certain frequencies will consistently sound louder to you. If the speed of sound is 343 m/s, what are the two lowest such frequencies? Waves of any given frequency are arriving at your ears via these two different paths, and they will sound especially loud to you if they are interfering constructively (i.e. adding their amplitudes to form a double amplitude). That will occur only if the path length difference is a whole number of wavelengths for that particular frequency: Δ x = n λ , where n = 0, 1, 2, 3... (In this case, we already know that Δ x ≠ 0, so that means n ≠ 0 here.) Of course, λ = v / f / , so the condition for constructive interference can also be written as Δ x = nv / f. Rearranging to solve for f , we get this : f = nv / Δ x Clearly, the two lowest possible frequencies would be for n -values of 1 and 2: f = (1) v / Δ x = 343/(5.50 – 3.86) = 209 Hz f = (2) v / Δ x = (2) 343/(5.50 – 3.86) = 418 Hz 2. A row of seats is parallel to a stage at a distance of 8.7 m from the stage. At the center and front of the stage is a loudspeaker with a circular aperture of diameter 7.5 cm. The speaker is playing a tone with a frequency of 10,000 Hz, and the speed of sound is 343 m/s. What is the shortest distance between any two seats located on opposite sides of the center of that row, at which this tone cannot be heard well? The characteristic dispersion angle, θ , gives the angle, speaker aperture as measured from the central axis (which extends out θ from the apertureʼs center, as shown), where diffracting sound waves will interfere destructively . L For a circular aperture of diameter r D : sin θ = 1.22 λ / D But of course: λ = v / f <-- seats --> Therefore: sin θ = 1.22 v /( fD ( ) And thus: θ = sin -1 [1.22 v /( fD ( )] The question is asking for the distance 2 x 2 2 : the minimum distance between two “damped out” seats on oppo- site sides of the center. Solving the right triangle shown for the distance x is now just a bit of trigonometry : x / L / = tan θ x = L tan θ 2 x = 2 L tan θ 2 x = 2 L tan{sin -1 [1.22 v /( fD ( )]} = 11.7 m
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Oregon State University Physics 202 , Winter 2010 HW5 W (due Feb. 8 at 5:00 p.m.) Page 2 3 . A certain string (fixed at both ends) will vibrate in a standing wave at a frequency of 113 Hz. The speed A of the waves along the string is 270 m/s. How far apart are the nodes? This question is not asking anything about which harmonic ( n -value) this wave is resonating at. We donʼt know that, nor do we need to know that—nor anything about the length, L , of the string. The nodes in any standing wave on a string (or in air, for that matter) occur every half-wavelength. That is, the distance between nodes is always λ /2. Since λ = v / f / , this distance is given by v /(2 f 2 2 ), which in this case would be 270/[2(113)] = 1.19 m .
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