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1 Quiz 1 solutions are in order of the questions for version A. (1). Correct answer A The lowest frequency for a string clamped at both ends is f = c/2L (2). Correct answer is B 345 This problem involves two Doppler shifts. First, the wall hear a sound fwall = f0 345−17 with f0 = 345/λ and then the reflected sound is heard with frequency f = fwall 345+17 . 345 Thus 345 ∆f ≡ f − f0 = .15 362 − 1 ' 240Hz 328 (3). Correct answer is A When the string creates a vibration in the air, they must have the same frequency (4). Correct answer is D Waves do not transport matter. All the other statements are true (5). Correct answer is A The frequency of organ pipe modes (open-open condition) is fn = n v 2L Hence the difference between any two successive modes is just 345/2L = 30Hz; solving for L gives answer A (6). Correct answer is A The path difference δ must equal half a wavelength. We get the correct answer from δ= p p 8.32 + (2.26 + 2.11)2 − 8.32 + (2.26 − 2.11)2 (7). Correct answer is A The period of time between intensity maxima is just 1/(beat frequency). Here the beat frequency is just the difference of the two frequencies, i.e. 38/min = 38/60 Hz (8). Correct answer is D The tension is just F= µv 2 , where µ is the mass per unit length. This can determined from the density by multiplying with the cross sectional area; hence F = 2700 ∗ π ∗ (2.3 × 10−3 )2 ∗ (30)2 ' 40N 2 (9). Correct answer is D Energy transport is always proportional to the square of the amplitude (10). Correct answer is D The fact that sound intensity goes down from he source is a direct consequence of conservation of energy